Lift — Kutta-Joukowski theorem L = ρV∞Γ

2.2.25 · Physics › Fluid Mechanics

WHAT — Pehle Definitions

WHY — $L^{'} = ρ V_{\infty} Γ$ aata kahan se hai?

Hum isse first principles se do complementary tareeqon se derive karte hain.

Derivation 1 — Bernoulli + ek model flow (physical picture)

Yeh kyun kaam karta hai: lift = net pressure force. Pressure differences speed differences se aati hain (Bernoulli). Circulation exactly wahi cheez hai jo top aur bottom ke beech speed difference create karti hai.

Airfoil ko flat plate ki tarah model karo freestream $V_{\infty}$ mein, jisme ek superimposed circulation hai jo top aur bottom par ek chhoti swirl velocity $v$ add karta hai.

Top speed: $V_{top} = V_{\infty} + v$
Bottom speed: $V_{bot} = V_{\infty} - v$

Kyun? Ek clockwise circulation $Γ$ top par freestream direction mein velocity add karta hai aur bottom par oppose karta hai.

Bernoulli (height change nahi): $p + \frac{1}{2} ρ V^{2} = const$ , toh $Δ p = p_{bot} - p_{top} = \frac{1}{2} ρ (V_{top}^{2} - V_{bot}^{2}) .$

Yeh step kyun? Zyada speed ⇒ kam pressure, isliye top-pressure expression ko bottom se minus karo.

$V_{top}^{2} - V_{bot}^{2} = (V_{\infty} + v)^{2} - (V_{\infty} - v)^{2} = 4 V_{\infty} v .$

Toh $Δ p = 2 ρ V_{\infty} v$ . Chord $c$ par lift per unit span hai $L^{'} = Δ p \cdot c = 2 ρ V_{\infty} (v c)$ .

Ab $v$ ko circulation se relate karo. Top par peeche ki taraf aur bottom par aage ki taraf extra swirl chord $c$ par contribute karta hai (rough loop estimate): $Γ = \oint V \cdot d l \approx (V_{\infty} + v) c - (V_{\infty} - v) c = 2 v c .$

Kyun? Loop ke dominant legs top aur bottom hain; freestream parts cancel ho jaate hain, $2 v c$ bachta hai.

$v c = Γ/2$ substitute karo: $L^{'} = 2 ρ V_{\infty} \cdot \frac{Γ}{2} = ρ V_{\infty} Γ. ✓$

Derivation 2 — Momentum / vector form (clean wala)

Uniform stream $V_{\infty}$ aur circulation magnitude $Γ$ ke liye, exact vector statement hai $L^{'} = ρ V_{\infty} \times Γ,$ jisme $∣ L^{'} ∣ = ρ V_{\infty} Γ$ , lift freestream ke perpendicular hai. Isliye airfoil ki lift incoming wind ke right angles par hoti hai, vertically upar nahi, aur yeh directly pressure integrate karne se aata hai (Blasius/complex-potential theorem se) body ke around — $ρ V_{\infty} Γ$ ke alawa har term integrate hokar zero ho jaata hai.

HOW — Use karna aur Kutta condition

Figure — Lift — Kutta-Joukowski theorem L = ρV∞Γ

Worked Examples

Common Mistakes (Steel-manned)

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho tum haath mein spinning top lekar daud rahe ho. Spin hawa ko pakad ke ghuma deta hai. Ab still hawa mein aage daudao: hawa ek side par tezi se jaati hai (jahan spin aur wind ek direction mein hain) aur doosri side par dheere. Tez hawa kam push karti hai, dheemi hawa zyada push karti hai — toh ball sideways dhakkel jaati hai. Ek wing wahi trick karta hai: uski shape hawa ko apne around ghuma deti hai, aur woh swirl ( $Γ$ ) times kitna tez udo ( $V$ ) times hawa kitni thick hai ( $ρ$ ) woh upar ka push deta hai jo plane ko upar rakkhta hai.

Active Recall

Kutta–Joukowski theorem batao aur har term ke units.

L^{'} = ρ V_{\infty} Γ

;

L^{'}

N/m mein,

ρ

kg/m³ mein,

V_{\infty}

m/s mein,

Γ

m²/s mein.

Circulation

Γ

define karo.

Γ = \oint_{C} V \cdot d l

, velocity ka closed-loop line integral; units m²/s.

Real airfoil ke liye

Γ

ki actual value kaun si physical condition select karti hai?

Kutta condition — flow ko sharp trailing edge se smoothly nikalna chahiye (finite velocity).

Airfoil ki shape

L^{'} = ρ V_{\infty} Γ

mein kyun nahi aati?

Kyunki saare shape-dependent pressure contributions body ke around integrate hokar zero ho jaate hain; sirf circulation term bachti hai (Blasius theorem).

Thin-airfoil circulation vs angle of attack?

Γ = π V_{\infty} c α

, jisse

c_{ℓ} = 2 π α

milta hai.

Lift freestream ke perpendicular kyun hoti hai, vertically upar nahi?

Vector form

L^{'} = ρ V_{\infty} \times Γ

se, force

V_{\infty}

ke right angles par hoti hai.

"Equal transit time" myth ka steel-man fix.

Hawa ko milna zaroori nahi; lift Kutta condition se set hone wale circulation se aati hai, path-length matching se nahi.

Ek ball wind mein spin kare. Kaun sa effect, wahi formula?

Magnus effect; sideways force

= ρ V_{\infty} Γ

jisme

Γ

rotation se aata hai.

Connections

Bernoulli's Principle — Derivation 1 mein use hone wala pressure–speed link deta hai.
Circulation and Vorticity — $Γ$ define karta hai aur ise Stokes' theorem ke zariye vorticity se relate karta hai.
Kutta Condition — unknown $Γ$ fix karta hai.
Magnus Effect — spinning bodies ke liye wahi theorem.
Thin Airfoil Theory — $Γ = π V_{\infty} c α$ deta hai.
Drag — d'Alembert's Paradox — kyun inviscid theory lift predict karta hai lekin zero drag.
Blasius Theorem — K–J ka rigorous complex-analysis proof.