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

Kya hai: K–J theorem mein seedha plug-in karna hai. Kyun: hume exactly teen quantities $\rho ,V_{\infty },\Gamma$ di gayi hain — aur kuch bhi zaroori nahi. $L^{\prime }=\rho V_{\infty }\Gamma =1.23\times 40\times 20=984\text{N/m}.$ Units check: $\frac{\text{kg}}{{\text{m}}^3}\cdot \frac{\text{m}}{\text{s}}\cdot \frac{{\text{m}}^2}{\text{s}}=\frac{\text{kg⋅m}}{{\text{s}}^2}\cdot \frac{1}{\text{m}}=\text{N/m}$ ✓

Kya hai: unknown ke liye K–J ko rearrange karna hai. Kyun: $L^{\prime }$ linearly $\Gamma$ par depend karta hai, isliye hum sirf divide karte hain. $\Gamma =\frac{L^{\prime }}{\rho V_{\infty }}=\frac{1500}{1.23\times 50}=\frac{1500}{61.5}=24.39{\text{m}}^2/\text{s}.$

• $L^{\prime }$: $\text{N/m}$ (force per metre of span)
• $\rho$: ${\text{kg/m}}^3$
• $V_{\infty }$: $\text{m/s}$
• $\Gamma$: ${\text{m}}^2/\text{s}$

Prime (${}^{\prime }$) ek per-unit-span quantity ko mark karta hai. Ek real wing ki total lift $L=L^{\prime }\times b$ hoti hai jahan $b$ span hai.

Step 1 — radians mein convert karo. Kyun: thin-airfoil formula $\Gamma =\pi V_{\infty }c\alpha$ radians mein $\alpha$ ke saath derive kiya gaya hai (yeh ek small-angle sine se aata hai). $\alpha =4^{\circ }\times \frac{\pi }{180}=0.06981\text{rad}.$ Step 2 — circulation. $\Gamma =\pi V_{\infty }c\alpha =\pi \times 60\times 1.5\times 0.06981=19.74{\text{m}}^2/\text{s}.$ Step 3 — lift. $L^{\prime }=\rho V_{\infty }\Gamma =1.23\times 60\times 19.74=1457\text{N/m}.$

(a) Thin-airfoil prediction: $c_{\ell }=2\pi \alpha =2\pi (0.06981)=\mathrm{0.4386.}$ (b) Computed lift se: $\frac{1}{2}\rho V_{\infty }^{2}c=\frac{1}{2}(1.23)(60^2)(1.5)=\frac{1}{2}(1.23)(3600)(1.5)=3321\text{N/m}.$ $c_{\ell }=\frac{L^{\prime }}{3321}=\frac{1457}{3321}=\mathrm{0.4387.}$ Dono match karte hain ✓ — thoda sa difference rounding ki wajah se hai. Yeh confirm karta hai ki $L^{\prime }=\rho V_{\infty }\Gamma$ aur $L^{\prime }=\frac{1}{2}\rho V_{\infty }^{2}c{c}_{\ell }$ ek hi statement ke do tarike hain.

Kya hai: Magnus Effect mein same theorem use hota hai — K–J ki parwah nahi ki $\Gamma$ wing shape se aaya ya spin se. $F^{\prime }=\rho V_{\infty }\Gamma =1.23\times 30\times 6=221.4\text{N/m}.$ Force ball ki velocity ke perpendicular hoti hai (isliye ball curve karti hai).

Key insight: fixed $\alpha$ aur $c$ par, circulation khud speed ke saath scale karta hai: $\Gamma =\pi V_{\infty }c\alpha \propto V_{\infty }$. Isliye do hidden speed factors hain. $L^{\prime }=\rho V_{\infty }\Gamma =\rho V_{\infty }(\pi V_{\infty }c\alpha )=\pi \rho c\alpha V_{\infty }^2\propto V_{\infty }^2.$ $V_{\infty }$ double karne se $L^{\prime }$ $2^2=\boxed{4}$ se multiply hota hai. Kyun sirf $\times 2$ nahi: theorem $V_{\infty }$ mein linear hai aur $\Gamma$ mein linear hai — lekin $\Gamma$ bhi speed ke saath badha, isliye dono multiply hote hain.

Wing A (clockwise, $\Gamma >0$): top flow speed up hoti hai, bottom slow down hoti hai $\Rightarrow$ top par low pressure $\Rightarrow$ lift upar jaati hai. Yeh normal lifting case hai. Wing B (counter-clockwise, $\Gamma <0$): sab kuch mirror hota hai — top slow hoti hai, bottom speed up hoti hai $\Rightarrow$ top par high pressure $\Rightarrow$ force neeche jaati hai (negative lift / down-force, jaise race car par ulti wing). Kyun sign cleanly flip hota hai: $L^{\prime }=\rho V_{\infty }\Gamma$ $\Gamma$ mein odd hai — swirl reverse karo, force reverse hoti hai. Magnitude $|L^{\prime }|=\rho V_{\infty }|\Gamma |$ identical hoti hai.

Step 1 — har factor kaise badalta hai.

• $\Gamma =\pi V_{\infty }c\alpha$: $\alpha$ aadha karna $\Gamma$ aadha kar deta hai, factor $\times \frac{1}{2}$.
• $L^{\prime }=\rho V_{\infty }\Gamma$: density aadhi ho jaati hai, factor $\times \frac{1}{2}$.
• $V_{\infty }$, $c$ unchanged. Step 2 — combine (dono $L^{\prime }$ par act karte hain): $\frac{L_{\text{new}}^{\prime }}{L_{\text{old}}^{\prime }}=\underset{\rho }{\underbrace{\frac{1}{2}}}\times \underset{\alpha \Rightarrow \Gamma }{\underbrace{\frac{1}{2}}}=\frac{1}{4}.$ Nayi lift original ki one-quarter hai.

(a) Lift per unit span — total lift weight ke equal honi chahiye, span $b$ par spread karke: $L^{\prime }=\frac{W}{b}=\frac{8000}{8}=1000\text{N/m}.$ (b) Circulation — K–J invert karo: $\Gamma =\frac{L^{\prime }}{\rho V_{\infty }}=\frac{1000}{1.23\times 45}=\frac{1000}{55.35}=18.07{\text{m}}^2/\text{s}.$ (c) Angle of attack — thin-airfoil relation $\Gamma =\pi V_{\infty }c\alpha$ invert karo: $\alpha =\frac{\Gamma }{\pi V_{\infty }c}=\frac{18.07}{\pi \times 45\times 1.4}=\frac{18.07}{197.9}=0.09130\text{rad}.$ $\alpha =0.09130\times \frac{180}{\pi }=5.231^{\circ }.$ Ideas ki chain jo use ki: weight → $L^{\prime }$ (per-span) → $\Gamma$ (K–J) → $\alpha$ (thin-airfoil). Har link ek formula hai.

Step 1: $c_{\ell }=2\pi \alpha =2\pi (0.09130)=0.5736$. Step 2: dynamic-pressure-times-chord: $\frac{1}{2}\rho V_{\infty }^{2}c=\frac{1}{2}(1.23)(45^2)(1.4)=\frac{1}{2}(1.23)(2025)(1.4)=1743.5\text{N/m}.$ Step 3: $L^{\prime }=1743.5\times 0.5736=1000\text{N/m}$ ✓ Kyun agree karna zaroori hai: $\frac{1}{2}\rho V_{\infty }^{2}c(2\pi \alpha )$ aur $\rho V_{\infty }(\pi V_{\infty }c\alpha )$ algebraically identical hain — dono $\pi \rho V_{\infty }^{2}c\alpha$ ke equal hain.

Kya hai: lift linearly $\alpha$ ke saath scale karti hai ($\Gamma$ ke through), isliye pichle angle ko lift ratio se multiply karo. Step 1: new $L^{\prime }=11200/8=1400\text{N/m}$. Step 2: $L^{\prime }\propto \alpha$ fixed $\rho ,V_{\infty },c$ par. Isliye ${\alpha }_{\text{new}}={\alpha }_{\text{old}}\times \frac{1400}{1000}=5.231^{\circ }\times 1.4=7.323^{\circ }.$ Sanity: abhi bhi ek small angle hai (< $10^{\circ }$), isliye thin-airfoil theory valid rehti hai.

Trap: student ne $\Gamma$ ko constant maana. Lekin fixed geometry par $\Gamma =\pi V_{\infty }c\alpha \propto V_{\infty }$, isliye $\Gamma$ bhi double hota hai jab speed double hoti hai. $L^{\prime }=\pi \rho c\alpha V_{\infty }^2.$

• Density $\times \frac{1}{2}$: factor $\frac{1}{2}$.
• Speed $\times 2$ inside $V_{\infty }^2$: factor $2^2=4$. $\frac{L_{\text{new}}^{\prime }}{L_{\text{old}}^{\prime }}=\frac{1}{2}\times 4=2.$ Student galat hai — lift double hoti hai, same nahi rehti.

Direction: ek cross product $\vec{A}\times \vec{B}$ hamesha $\vec{A}$ aur $\vec{B}$ dono ke perpendicular hota hai. Kyunki ${\vec{L}}^{\prime }=\rho {\vec{V}}_{\infty }\times \vec{\Gamma }$, lift ${\vec{V}}_{\infty }$ ke perpendicular hai — yaani wind ke right angles par, ground ke nahi. Magnitude (freestream aur circulation-axis perpendicular hain, isliye $\sin 90^{\circ }=1$): $|L^{\prime }|=\rho V_{\infty }\Gamma =1.20\times 70\times 22=1848\text{N/m}.$ d'Alembert: ideal (inviscid, incompressible) 2-D flow mein ${\vec{V}}_{\infty }$ ke along net force — drag — exactly zero hai. Pura K–J perpendicular component mein hai. Real drag ke liye viscosity chahiye, jo yeh theory omit karti hai.

(a) Myth ki top speed — same travel time, longer path: $V_{\text{top}}=V_{\infty }\times \frac{1.05c}{c}=50\times 1.05=52.5\text{m/s}.$ (b) Bernoulli pressure difference (koi height change nahi): $\Delta p=\frac{1}{2}\rho \left(V_{\text{top}}^2-V_{\infty }^2\right)=\frac{1}{2}(1.23)(52.5^2-50^2).$ $52.5^2-50^2=2756.25-2500=\mathrm{256.25.}$ $\Delta p=\frac{1}{2}(1.23)(256.25)=157.6\text{Pa},L^{\prime }=\Delta p\times c=157.6\text{N/m}.$ (c) Lesson: is size ki real wings $157.6\text{N/m}$ se das-plus guuna zyada lift karti hain. Isliye equal-transit assumption lift ko bahut bura under-predict karti hai — proving ki air dobara nahi milti (top air actually pehle pahunchti hai). Lift Kutta Condition ke through selected circulation se set hoti hai, phir $L^{\prime }=\rho V_{\infty }\Gamma$ se deliver hoti hai, kisi rejoining rule se nahi.

Kya number $\text{N/m}$ mein aaya? ::: Haan agar tumne $\rho V_{\infty }\Gamma$ use kiya aur span se kabhi multiply nahi kiya. Kya $\Gamma =\pi V_{\infty }c\alpha$ mein daalne se pehle har angle radians mein tha? ::: Pehle $\times \pi /180$ se convert karo. Agar geometry fixed thi, kya $\Gamma$ $V_{\infty }$ ke saath saath chala? ::: Tab $L^{\prime }\propto V_{\infty }^2$ hai, $V_{\infty }$ nahi. Kya tumne span se exactly ek baar divide kiya (sirf total $L$ se $L^{\prime }$ paane ke liye)? ::: Ek division, na zyada na kam.