Exercises — Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c)

Theorem yeh hai: $L^{\prime }=\rho V\Gamma .$

• $L^{\prime }$ — lift per unit span, units $\text{N/m}$.
• $\rho$ — density, ${\text{kg/m}}^3$.
• $V$ — speed, $\text{m/s}$.
• $\Gamma$ — circulation, ${\text{m}}^2/\text{s}$.

Check karo ki units sahi multiply ho rahi hain: $\frac{\text{kg}}{{\text{m}}^3}\cdot \frac{\text{m}}{\text{s}}\cdot \frac{{\text{m}}^2}{\text{s}}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}\cdot \frac{1}{\text{m}}=\frac{\text{N}}{\text{m}}.$ ✓ Lift span ke har metre ke liye hai, isliye prime kabhi nahi hatana chahiye. Dekho Kutta–Joukowski theorem.

$c_{\ell }=2\pi \alpha$ ko $\alpha$ ke saath differentiate karo: $\frac{d{c}_{\ell }}{d\alpha }=2\pi \approx 6.283\text{ per radian.}$ "Per radian" matter karta hai: $\alpha$ radians mein hona chahiye kyunki derivation ne $\sin \alpha$ ko $\alpha$ se replace kiya, yeh approximation sirf radians mein measure kiye gaye chote angles ke liye valid hai.

Step 1 — angle convert karo. $\alpha =4^{\circ }=4\times \frac{\pi }{180}=0.06981$ rad. Kyun: formula radians assume karta hai.

Step 2 — camber term zero hai. $\beta =0$, isliye $\alpha +\frac{2\beta }{\pi }=0.06981$.

Step 3 — lift formula mein plug karo. $L^{\prime }=\pi \rho V^{2}c\left(\alpha +\frac{2\beta }{\pi }\right)=\pi (1.225)(60^2)(1.2)(0.06981).$ $L^{\prime }=\pi (1.225)(3600)(1.2)(0.06981)\approx 1161\text{N/m}.$ Toh span ka har metre lagbhag $1161$ N carry karta hai.

Step 1 — camber contribution. $\frac{2\beta }{\pi }=\frac{2(0.04)}{\pi }=0.02546$.

Step 2 — total effective angle. $\alpha +\frac{2\beta }{\pi }=0.06981+0.02546=0.09527$.

Step 3 — lift. $L^{\prime }=\pi (1.225)(3600)(1.2)(0.09527)\approx 1585\text{N/m}.$

Step 4 — percentage increase. $\frac{1585-1161}{1161}\times 100\%\approx 36.5\%.$ Camber ne same angle of attack par $\sim 37\%$ lift add kar di — yeh poori curve ko upar shift kar deta hai. Dekho Lift coefficient and angle of attack.

(a) $c_{\ell }=2\pi (0.06981)+4(0.04)=0.4386+0.16=\mathrm{0.5986.}$

(b) Dynamic pressure part: $\frac{1}{2}\rho V^{2}c=0.5(1.225)(3600)(1.2)=2646$. $c_{\ell }=\frac{L^{\prime }}{2646}=\frac{1585}{2646}=\mathrm{0.599.}$ Dono $\approx 0.599$ dete hain. ✓ Yeh agree karte hain kyunki $c_{\ell }=2\pi \left(\alpha +\frac{2\beta }{\pi }\right)=2\pi \alpha +4\beta$ exactly wahi lift formula hai jo dynamic-pressure term se divide kiya gaya hai.

Step 1 — $L^{\prime }=0$ set karo. Lift tab khatam hoti hai jab bracket khatam ho: $\alpha +\frac{2\beta }{\pi }=0\Rightarrow {\alpha }_{L=0}=-\frac{2\beta }{\pi }.$ Yeh step kyun: $\pi \rho V^{2}c>0$ hamesha hota hai, isliye $L^{\prime }=0$ karne ka ek hi tarika hai — bracket ko khatam karo.

Step 2 — plug in karo. ${\alpha }_{L=0}=-\frac{2(0.04)}{\pi }=-0.02546$ rad.

Step 3 — degrees mein. $-0.02546\times \frac{180}{\pi }=-1.459^{\circ }.$

Physical matlab: ek cambered wing flow ke saath perfectly aligned hone par bhi hawa ko neeche scoop karta hai. Zero net downward turning produce karne ke liye tumhe nose ko thoda neeche tilt karna padta hai, isliye negative angle aata hai. Neeche figure mein shifted curve dekho — yeh origin ke baaye taraf zero cross karta hai.

Step 1 — yaad karo kaun se coefficients bachte hain. Parent derivation se, $\Gamma =\pi cV\left(A_0+\frac{A_1}{2}\right).$ Sirf $A_0$ aur $A_1$ enter karte hain, kyunki chord ke saath sheet ko integrate karne mein $\sin n\theta$ terms ki orthogonality use hoti hai; $n\ge 2$ wala har $A_n$, $[0,\pi ]$ par integrate ho kar zero deta hai. Dekho Glauert's integral and Fourier coefficients.

Step 2 — plug in karo. $\Gamma =\pi (1.0)(40)\left(0.09+\frac{0.05}{2}\right)=\pi (40)(0.115)=14.45{\text{m}}^2/\text{s}.$

Step 3 — $A_2$ lift ko kyun nahi dikhta. $A_2$ (aur usse upar wale) chord ke saath vorticity redistribute karte hain — yeh moment (pitching) ko change karte hain lekin total circulation mein, aur isliye total lift mein unka net contribution cancel ho jaata hai. Isliye $A_2=0.03$ yahan simply irrelevant hai.

Route A — Kutta–Joukowski. $L^{\prime }=\rho V\Gamma =(1.225)(40)(14.451)=708.1\text{N/m}.$

Route B — direct formula. $L^{\prime }=\pi \rho c{V}^2\left(A_0+\frac{A_1}{2}\right)=\pi (1.225)(1.0)(1600)(0.115)=708.1\text{N/m}.$ Dono routes match karte hain, kyunki Route B aslaan Route A hi hai jisme $\Gamma =\pi cV(A_0+A_1/2)$ substitute kiya gaya hai.

Step 1 — variable change karo. $x=\frac{c}{2}(1-\cos \theta )$ ke saath, $1-\frac{2x}{c}=1-(1-\cos \theta )=\cos \theta ,\text{so}\frac{dz}{dx}=m\cos \theta .$ Kyun: poora Glauert machinery $\theta$ variable mein rehta hai, isliye slope ko $\theta$ mein rewrite karna zaroori hai.

Step 2 — $A_0$ compute karo. $A_0=\alpha -\frac{1}{\pi }{\int }_0^{\pi }\frac{dz}{dx}d\theta =\alpha -\frac{1}{\pi }{\int }_0^{\pi }m\cos \theta d\theta .$ Kyunki ${\int }_0^{\pi }\cos \theta d\theta =[\sin \theta {]}_0^{\pi }=0$, humein milta hai $A_0=\alpha .$

Step 3 — $A_1$ compute karo. $A_1=\frac{2}{\pi }{\int }_0^{\pi }\frac{dz}{dx}\cos \theta d\theta =\frac{2m}{\pi }{\int }_0^{\pi }{\cos }^2\theta d\theta =\frac{2m}{\pi }\cdot \frac{\pi }{2}=m.$ (Yahan use kiya ${\int }_0^{\pi }{\cos }^2\theta d\theta =\pi /2$.)

Step 4 — lift bracket assemble karo. $A_0+\frac{A_1}{2}=\alpha +\frac{m}{2}.$ $\alpha +\frac{2\beta }{\pi }$ se compare karne par milta hai: $\boxed{\frac{2\beta }{\pi }=\frac{m}{2}\Rightarrow \beta =\frac{\pi m}{4}.}$ Toh abstract parameter $\beta$ bas maximum camber-line slope $m$ ka $\frac{\pi }{4}$ guna hai.

Step 1 — angle radians mein. $\alpha =3^{\circ }=0.05236$ rad.

Step 2 — lift bracket. $A_0+A_1/2=\alpha +m/2=0.05236+0.05=\mathrm{0.10236.}$

Step 3 — lift coefficient. $c_{\ell }=2\pi (A_0+A_1/2)=2\pi (0.10236)=\mathrm{0.6431.}$

Step 4 — lift per span. $L^{\prime }=\pi \rho V^{2}c(A_0+A_1/2)=\pi (1.225)(55^2)(1.4)(0.10236).$ $L^{\prime }=\pi (1.225)(3025)(1.4)(0.10236)\approx 1669\text{N/m}.$

Step 1 — total = per-span × span. $L=L^{\prime }\times b=1669\times 8=13352\text{N}\approx 13.4\text{kN}.$ Kyun: $L^{\prime }$ ki units N/m hain; span (m) se multiply karne par newtons milte hain. Isliye prime itna important tha poori time.

Model. Thin airfoil theory kehti hai $c_{\ell }=2\pi \left(\alpha +\frac{2\beta }{\pi }\right)$, yeh $\alpha$ mein ek straight line hai. Do points line fix kar dete hain.

Step 1 — angles convert karo. ${\alpha }_1=-2^{\circ }=-0.03491$ rad (jahan $c_{\ell }=0$); ${\alpha }_2=3^{\circ }=0.05236$ rad (jahan $c_{\ell }=0.55$).

Step 2 — do points se slope. $\text{slope}=\frac{c_{\ell ,2}-c_{\ell ,1}}{{\alpha }_2-{\alpha }_1}=\frac{0.55-0}{0.05236-(-0.03491)}=\frac{0.55}{0.08727}=6.303\text{per rad}.$ Theory predict karti hai $2\pi =6.283$. Measure kiya hua $6.303$ sirf $0.3\%$ andar hai — shandar agreement, confirm karta hai ki slope sach mein $\approx 2\pi$ hai aur camber ne use nahi badla.

Step 3 — zero-lift point se camber. Jab $c_{\ell }=0$ ho, $\alpha ={\alpha }_{L=0}=-\frac{2\beta }{\pi }$. $\frac{2\beta }{\pi }=-{\alpha }_{L=0}=-(-0.03491)=\mathrm{0.03491.}$ Toh effective camber $0.0349$ rad ka "extra angle" contribute karta hai, jo $\beta =\frac{\pi }{2}(0.03491)=0.05483$ rad ke barabar hai.

Step 1 — incompressible value (5.1 se): $c_{\ell ,\text{incomp}}=0.55$.

Step 2 — correction factor. $\sqrt{1-M_{\infty }^2}=\sqrt{1-0.36}=\sqrt{0.64}=\mathrm{0.8.}$ Yeh tool kyun: zyada speed par hawa compress hoti hai, wing ke paas zyada pressure difference pack hoti hai; Prandtl–Glauert incompressible answer ko $1/\sqrt{1-M^2}$ se scale karta hai. Yeh jawab deta hai "compressibility meri low-speed lift ko kitna amplify karti hai?"

Step 3 — corrected coefficient. $c_{\ell ,\text{comp}}=\frac{0.55}{0.8}=\mathrm{0.6875.}$ Compressibility ne lift coefficient $25\%$ badha di. Dekho Compressibility corrections (Prandtl–Glauert).

Step 1 — target likho. $c_{\ell }=2\pi \alpha +4\beta$ aur (Problem 4.1) $4\beta =\pi m$, isliye equivalently $c_{\ell }=2\pi \alpha +\pi m$. Kyun: sab kuch design knob $m$ mein express karo.

Step 2 — angle radians mein. $\alpha =2^{\circ }=0.03491$ rad.

Step 3 — $m$ ke liye solve karo. $0.80=2\pi (0.03491)+\pi m\Rightarrow 0.80=0.21932+\pi m.$ $\pi m=0.58068\Rightarrow m=\frac{0.58068}{\pi }=\mathrm{0.1848.}$ Toh mean line ko lagbhag $m\approx 0.185$ ka maximum slope chahiye. Yeh kaafi aggressive camber hai — sanity check: yeh zero-lift angle ${\alpha }_{L=0}=-m/2=-0.0924$ rad $=-5.3^{\circ }$ par rakhta hai, strong camber ke liye expected ke anusaar comfortably negative hai.

Lift bracket hai ==$\alpha +\frac{2\beta }{\pi }$ aur parabolic camber line ke liye $\alpha +\frac{m}{2}$ ke barabar hai. Circulation integral mein sirf $A_0$ aur $A_1$ bachte hain. Lift-curve slope hai $2\pi$ per radian aur camber sirf curve ko shift karta hai, slope ko nahi==. Prandtl–Glauert $c_{\ell }$ ko ==$1/\sqrt{1-M^2}$== se multiply karta hai.