Lift — Kutta-Joukowski theorem L = ρV∞Γ: Worked examples

2.2.25 · D3 · Physics › Fluid Mechanics › Lift — Kutta-Joukowski theorem L = ρV∞Γ

Yeh page parent topic ka drill hall hai. Hum ek single formula $L^{'} = ρ V_{\infty} Γ$ ko har us case mein push karte hain jo tumhare saamne aa sakta hai — positive aur negative circulation, zero inputs, degenerate flows, limiting behaviour, ek real-world word problem, aur ek exam twist. Iske baad tumhe koi bhi aisa scenario nahi milna chahiye jo tumne pehle worked out na dekha ho.

Scenario matrix

Is theorem se aane wala har problem inhi cells mein se kisi ek mein land karta hai. Har worked example us cell ke saath tagged hai jise woh cover karta hai.

#	Cell class	Kya test ho raha hai	Worked in
A	Forward, $Γ > 0$	seedha plug-in, upward lift	Ex 1
B	Sign flip, $Γ < 0$	negative/downward lift, direction	Ex 2
C	Zero input	$Γ = 0$ ya $V_{\infty} = 0$ → zero lift (d'Alembert)	Ex 3
D	$Γ$ from geometry	thin-airfoil $Γ = π V_{\infty} c α$	Ex 4
E	Limiting/scaling	jab $V_{\infty}$ double ho toh $L^{'}$ kaise scale karta hai	Ex 5
F	Real-world word problem	total lift over a span, back-solve $Γ$	Ex 6
G	Non-airfoil source ( $Γ$ from spin)	Magnus, sideways force	Ex 7
H	Exam twist / vector direction	$L^{'} = ρ V_{\infty} \times Γ$ , kya lift vertical hai?	Ex 8

Poori note mein yeh constants use honge: sea-level air $ρ = 1.23 kg/m^{3}$ jab tak kuch aur na bataya jaye.

Cell A — Forward flight, positive circulation

Cell B — Negative circulation, downward force

Worked example Example 2 — swirl palto, lift palto

Ex 1 jaisi hi air aur speed ( $ρ = 1.23$ , $V_{\infty} = 40 m/s$ ), lekin ab flow counter-clockwise swirl kar raha hai: $Γ = - 25 m^{2} / s$ . $L^{'}$ kya hai aur yeh kis direction mein point karta hai?

Forecast: number Ex 1 jaisa hi size ka hona chahiye — lekin sign kaunsa?

Step 1. Same theorem, $L^{'} = ρ V_{\infty} Γ$ . Yeh step kyun? Theorem $Γ$ mein linear hai; $Γ$ ka sign flip karne se $L^{'}$ ka sign flip hota hai aur kuch nahi.

Step 2. $L^{'} = 1.23 \times 40 \times (- 25) = - 1230 N/m$ . Yeh step kyun? Negative result ka matlab hai force positive- $Γ$ convention ki "up" se opposite direction mein point karta hai — yaani yahan downward (yeh exactly usse-down wing ka case hai, ya inverted race-car wing jo downforce produce karta hai).

Step 3. Magnitude aur direction report karo: $∣ L^{'} ∣ = 1230 N/m$ , neeche point kar raha hai.

Verify: $L^{'} (Γ) = - L^{'} (- Γ)$ , toh Ex 1 aur Ex 2 ki magnitude equal hai, sign opposite ✓. Race-car front wings sach mein isi cheez ka use karte hain tyres ko track pe press karne ke liye.

Cell C — Zero inputs aur d'Alembert's paradox

Worked example Example 3 — zero lift paane ke do tarike

(a) Ek perfectly streamlined shape ideal flow mein koi circulation generate nahi karti: $Γ = 0$ . (b) Ek wind tunnel mein baithe wing ka tunnel band ho jaata hai: $V_{\infty} = 0$ . Dono cases mein $L^{'}$ nikalo.

Forecast: dono zero denge — lekin reasons alag hain. Kya tum pehle bata sakte ho kyun?

Step 1 (case a). $L^{'} = ρ V_{\infty} \cdot 0 = 0$ . Yeh step kyun? Jab koi swirl nahi, toh top/bottom speed difference nahi, toh Bernoulli pressure difference nahi deta. Yeh exactly d'Alembert's paradox hai: circulation ke bina ideal flow na drag produce karta hai na lift.

Step 2 (case b). $L^{'} = ρ \cdot 0 \cdot Γ = 0$ . Yeh step kyun? Koi oncoming momentum nahi jo deflect ho sake. Circulation akele (ek stationary vortex) wing ke paast kuch transport nahi karta, toh freestream deflection se koi net force nahi. Lift ke liye tumhe udna padega tabhi swirl kaam aayega.

Step 3. Dono cases: $L^{'} = 0$ .

Verify: Formula teen factors ka product hai; agar koi bhi factor $0$ hai, product $0$ hai ✓. Physically alag causes, identical zero result.

Cell D — Circulation forced by geometry (thin airfoil)

Worked example Example 4 — tumhe shape di gayi hai,

$Γ$ nahi Ek thin airfoil ki chord $c = 1.5 m$ hai, $V_{\infty} = 55 m/s$ pe ud raha hai, angle of attack $α = 4^{\circ}$ , air $ρ = 1.23$ . Pehle $Γ$ nikalo, phir $L^{'}$ .

Forecast: yahan $Γ$ seedha nahi diya — Kutta condition ise fix karti hai. Guess karo $L^{'}$ hazar ke kareeb hoga ya kaafi hazaron N/m?

Step 1. Angle convert karo: $α = 4^{\circ} = 4 \times \frac{π}{180} = 0.0698 rad$ . Yeh step kyun? Thin-airfoil formula $Γ = π V_{\infty} c α$ small-angle calculus se bana hai, isliye $α$ radians mein hona chahiye (degrees galat hoga by a factor $180/ π \approx 57$ ).

Step 2. $Γ = π V_{\infty} c α = π \times 55 \times 1.5 \times 0.0698 = 18.09 m^{2} / s$ . Yeh step kyun? Kutta condition (sharp trailing edge ke paast smooth flow) exactly yahi $Γ$ pick karti hai — na zyada, na kam.

Step 3. $L^{'} = ρ V_{\infty} Γ = 1.23 \times 55 \times 18.09 = 1224 N/m$ .

Verify: Lift coefficient se cross-check karo. Thin-airfoil kehta hai $c_{ℓ} = 2 π α = 2 π (0.0698) = 0.4386$ . Phir $L^{'} = \frac{1}{2} ρ V_{\infty}^{2} c c_{ℓ} = 0.5 \times 1.23 \times 5 5^{2} \times 1.5 \times 0.4386 = 1224 N/m$ ✓ — dono roads milti hain.

Cell E — Scaling aur limiting behaviour

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

Worked example Example 5 — speed double karne pe kya hota hai?

Ek wing $V_{\infty} = 55 m/s$ pe $L^{'} = 1224 N/m$ deta hai (Ex 4 se, same airfoil aur $α$ ). Ab tum $V_{\infty} = 110 m/s$ pe ud rahe ho. Naya $L^{'}$ predict karo bina har number re-plug kiye.

Forecast: K–J theorem $V_{\infty}$ mein linear hai — toh kya answer bas $2 \times$ hai? Figure dekhne ke baad decide karo.

Step 1. Note karo ki fixed shape aur angle ke liye, $Γ = π V_{\infty} c α \propto V_{\infty}$ . Yeh step kyun? Circulation khud speed ke saath badhti hai (faster flow same trailing edge se zyada swirl karta hai). Yeh woh "hidden second $V_{\infty}$ " hai jiske baare mein parent note ne warn kiya tha.

Step 2. Toh $L^{'} = ρ V_{\infty} Γ \propto V_{\infty} \times V_{\infty} = V_{\infty}^{2}$ . Yeh step kyun? Ek $V_{\infty}$ factor explicit hai, ek $Γ$ mein chhupta hai. Multiply karne pe $L^{'}$ speed ke square ke according scale karta hai — chahe theorem har factor mein alag se linear ho.

Step 3. Speed double → $L^{'}$ $2^{2} = 4$ se badhta hai: naya $L^{'} = 4 \times 1224 = 4896 N/m$ .

Verify: Direct compute karo. Naya $Γ = π \times 110 \times 1.5 \times 0.0698 = 36.17 m^{2} / s$ ; $L^{'} = 1.23 \times 110 \times 36.17 = 4894 N/m$ (rounding) ✓. Figure mein red $V_{\infty}^{2}$ curve wohi hai jo tum takeoff pe feel karte ho: thoda sa speed gain, bada lift gain.

Cell F — Real-world word problem (back-solve $Γ$ )

Worked example Example 6 — ek light plane ko kitna swirl upar rakhta hai?

Ek chhote aircraft ki mass $m = 1200 kg$ hai, wingspan $b = 11 m$ , $V_{\infty} = 60 m/s$ pe cruise karta hai, air $ρ = 1.23$ . Maano lift span ke along uniform hai, toh wing ke har metre mein kitna circulation $Γ$ hai?

Forecast: total lift weight ke barabar hona chahiye. Pehle weight newtons mein estimate karo ( $\approx$ 12 000 N), phir $Γ$ guess karo.

Step 1. Support karne wala weight: $W = m g = 1200 \times 9.81 = 11772 N$ . Yeh step kyun? Steady level flight mein, total lift $L = W$ . Yeh poore problem ko ek real force se anchor karta hai.

Step 2. Zaroori lift per unit span: $L^{'} = \frac{L}{b} = \frac{11 772}{11} = 1070.2 N/m$ . Yeh step kyun? K–J $L^{'}$ (N/m) deta hai, isliye total lift ko pehle per-span mein convert karna hoga — classic "per unit span" trap.

Step 3. Theorem ko $Γ$ ke liye invert karo: $Γ = \frac{L ^{'}}{ρ V _{\infty}} = \frac{1070.2}{1.23 \times 60} = 14.50 m^{2} / s$ . Yeh step kyun? Humein $L^{'}$ , $ρ$ , $V_{\infty}$ pata hai; sirf $Γ$ unknown hai, toh algebraically solve karo.

Verify: Forward rebuild karo: $L^{'} = 1.23 \times 60 \times 14.50 = 1070 N/m$ ; $\times b = 1070 \times 11 = 11770 N \approx W$ ✓. Lagbhag $14.5 m^{2} / s$ swirl per metre plane ko hawaai rakhta hai.

Cell G — Circulation from spin (Magnus effect)

Worked example Example 7 — curving football

Ek football, jiska koi aerofoil shape nahi hai, $V_{\infty} = 25 m/s$ pe kick kiya gaya hai jisme heavy spin hai jo air ko $Γ = 8 m^{2} / s$ circulation mein kheench leta hai. Sideways force per unit length nikalo.

Forecast: ball gol hai, wing-shaped nahi — kya K–J phir bhi apply hoti hai? (Hoti hai.)

Step 1. Same theorem apply karo: $F^{'} = ρ V_{\infty} Γ$ . Yeh step kyun? K–J ko sirf $ρ$ , $V_{\infty}$ , $Γ$ se matlab hai — isse nahi ki swirl shape se aaya ya spin se. Yeh Magnus effect hai, wing lift ka ek seedha cousin.

Step 2. $F^{'} = 1.23 \times 25 \times 8 = 246 N/m$ . Yeh step kyun? Same product structure; sirf $Γ$ ka source alag hai.

Step 3. Direction: flight path ke perpendicular, us side mein jahan spin aur airflow agree karte hain — yahi curved free kick ka "bend" hai.

Verify: Units → N/m ✓. Spin half karna ( $Γ \to 4$ ) curve ko half karke $123 N/m$ kar dega ✓, jo everyday intuition se match karta hai ki zyada spin = zyada bend.

Cell H — Exam twist: kya lift sach mein vertical hai?

Worked example Example 8 — magnitude nahi, direction

Ek wing freestream $V_{\infty}$ ke saath horizontally $30 m/s$ pe ud raha hai, circulation $Γ = 20 m^{2} / s$ , air $ρ = 1.23$ . Exam poochh raha hai: (i) $L^{'}$ ki magnitude, (ii) kya lift seedhi upar hai, aur (iii) kya hoga agar freestream $1 0^{\circ}$ nose-down tilted aaye?

Forecast: hum sab "jaante" hain lift vertical hoti hai — lekin kya yeh ground ke relative define hoti hai ya wind ke?

Step 1. Magnitude: $L^{'} = ρ V_{\infty} Γ = 1.23 \times 30 \times 20 = 738 N/m$ . Yeh step kyun? Scalar form orientation se independent magnitude deta hai.

Step 2. Vector form $L^{'} = ρ V_{\infty} \times Γ$ se direction. Yeh step kyun? Cross product ka matlab hai lift hamesha $V_{\infty}$ ke perpendicular hoti hai, ground ke nahi.

Step 3. (ii) Horizontal freestream ke saath, "wind ke perpendicular" hi seedha upar hai, toh lift vertical hai — yahan yeh geometry ki coincidence ki wajah se sach hai. (iii) Agar freestream $1 0^{\circ}$ nose-down tilti hai, toh lift bhi perpendicular rehne ke liye usi ke saath tilti hai: yeh vertical se $1 0^{\circ}$ door point karta hai. Magnitude $738 N/m$ unchanged rehti hai.

Verify: Cross-product magnitude $∣ ρ V_{\infty} \times Γ ∣ = ρ V_{\infty} Γ$ (2-D mein yeh perpendicular hain), $= 738 N/m$ ✓. Figure dekho: $V_{\infty}$ ko rotate karne se lift arrow same angle se rotate hota hai, hamesha $9 0^{\circ}$ apart.

Active Recall

Recall Agar

$Γ$ ka sign flip ho, toh $L^{'}$ ka kya hota hai? Question ::: Ek wing clockwise $Γ$ ke saath $+ 1230 N/m$ deta hai. Counter-clockwise $- Γ$ kya dega? Answer ::: $- 1230 N/m$ — same magnitude, downward (downforce). $L^{'}$ , $Γ$ mein linear hai.

Recall Zero lift paane ke do tarike?

Question ::: Do independent inputs batao jo dono $L^{'} = 0$ force karte hain. Answer ::: $Γ = 0$ (koi swirl nahi — d'Alembert's paradox) ya $V_{\infty} = 0$ (koi flight nahi, koi momentum deflect karne ko nahi).

Recall Speed double karne pe lift kyun quadruple hoti hai fixed wing ke liye?

Question ::: Theorem $V_{\infty}$ mein linear hai; toh observed lift $\propto V_{\infty}^{2}$ kyun hai? Answer ::: Kyunki $Γ = π V_{\infty} c α$ bhi $V_{\infty}$ ke saath badhta hai, toh $L^{'} = ρ V_{\infty} Γ \propto V_{\infty}^{2}$ .

Recall Level flight se

$Γ$ back-solve karna? Question ::: Total lift weight $W$ ke barabar, span $b$ . $Γ$ ka formula? Answer ::: $Γ = \frac{W}{ρ V _{\infty} b}$ ( $L^{'} = W / b = ρ V_{\infty} Γ$ se).