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

3.1.21 · Physics › Compressible Flow & Aerodynamics

HUM KYA compute kar rahe hain?

Prime ( $L^{'}$ ) ka matlab hai per unit span — units N/m hain, N nahi. Hum ise derive karenge, kabhi directly nahi denge.

EK WING LIFT KYON KARTA HAI? (first principles)

HUM ISSE KAISE derive karte hain — the vortex sheet

Step 1 — Airfoil ko ek vortex sheet se replace karo. Chord ke saath ( $0 \leq x \leq c$ ) ek vorticity sheet $γ (x)$ rakh do. Har element ek tiny vortex ki tarah kaam karta hai jiska strength $γ d x$ hai.

Ye step kyun? Ek thin airfoil flow ko bahut kam disturb karta hai, isliye camber line khud vorticity carry kar sakti hai. Hume real curved surface ki zaroorat nahi — ek line of vortices waisa hi outer flow reproduce karta hai.

Step 2 — Flow-tangency (boundary) condition. Flow camber line ke tangent hona chahiye. Mean line ke normal par velocity (a) freestream aur (b) vortex sheet se cancel honi chahiye: $V (α - \frac{d z}{d x}) = w (x)$ jahan $z (x)$ camber shape hai aur $w (x)$ sheet se induce hua downwash hai.

Ye step kyun? Hawa surface ke through flow nahi kar sakti. Ye condition hi $γ (x)$ ko pin down karti hai.

Step 3 — Biot–Savart se Downwash. Position $ξ$ par ek vortex element $γ (ξ) d ξ$ , $x$ par ye induce karta hai: $w (x) = - \frac{1}{2 π} \int_{0}^{c} \frac{γ ( ξ ) d ξ}{x - ξ}$

Ye step kyun? Ye bas 2-D vortex velocity $v = Γ/2 π r$ hai jo sheet par sum ho raha hai.

Step 4 — Cosine transform substitute karo. Maan lo $x = \frac{c}{2} (1 - cos θ)$ . Standard Glauert solution hai $γ (θ) = 2 V (A_{0} \frac{1 + c o s θ}{s i n θ} + \sum_{n = 1}^{\infty} A_{n} sin n θ)$ jo Kutta condition satisfy karta hai (trailing edge par finite $γ$ ). Boundary condition mein plug karke aur Glauert's integral $\int_{0}^{π} \frac{c o s n θ ^{'}}{c o s θ ^{'} - c o s θ} d θ^{'} = \frac{π s i n n θ}{s i n θ}$ use karke Fourier coefficients milte hain $A_{0} = α - \frac{1}{π} \int_{0}^{π} \frac{d z}{d x} d θ, A_{n} = \frac{2}{π} \int_{0}^{π} \frac{d z}{d x} cos n θ d θ$

Step 5 — Total circulation aur lift. $Γ = \int_{0}^{c} γ d x = c V π (A_{0} + \frac{A _{1}}{2})$ $L^{'} = ρ V Γ = π ρ c V^{2} (A_{0} + \frac{A _{1}}{2})$

Ye step kyun? Sirf $A_{0}$ aur $A_{1}$ hi $Γ$ ke integration mein bachte hain ( $sin n θ$ ki orthogonality $n \geq 2$ ko khatam kar deti hai). Ye general result hai.

Step 6 — Simple cambered case. Ek symmetric flat plate ke liye, $d z / d x = 0 \Rightarrow A_{0} = α, A_{1} = 0$ , jisse ye famous result milta hai $L^{'} = π ρ c V^{2} α \Rightarrow c_{ℓ} = 2 π α .$ Parabolic camber line ke liye jiska combined camber contribution $A_{0} + A_{1} /2 = α + \frac{2 β}{π}$ ke barabar hai, hum target result recover karte hain: $L^{'} = π ρ V^{2} c (α + \frac{2 β}{π})$

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

Worked examples

Flashcards

Kutta–Joukowski theorem for 2-D lift

L^{'} = ρ V Γ

L^{'}

mein prime ka kya matlab hai?

Lift per unit span (N/m), total lift nahi.

Thin airfoil theory se lift-curve slope

d c_{ℓ} / d α = 2 π

per radian.

Camber ke saath section lift coefficient

c_{ℓ} = 2 π α + 4 β

Sharp airfoil par circulation ki magnitude kya set karta hai?

Kutta condition (trailing edge se smooth flow).

Lift integral mein kaun se Fourier coefficients bachte hain?

Sirf

A_{0}

aur

A_{1}

Γ = π c V (A_{0} + A_{1} /2)

Zero-lift angle of attack

α_{L = 0} = - 2 β / π

α

radians mein kyun hona chahiye?

Small-angle linearisation

sin α \approx α

ne radians assume kiye the.

Model mein airfoil ki jagah kya aata hai?

Camber line ke saath ek vortex sheet

γ (x)

Kya camber lift-curve slope change karta hai?

Nahi — ye sirf curve ko shift karta hai (constant offset).

Recall Feynman: 12-saal ke bachche ko samjhao

Frisbee phenkte waqt socho. Hawa ko upar ki taraf push karwane ke liye, wing ko hawa ko neeche dhakhelna hota hai. Ek flat tilti card hawa ko thoda neeche dhakhelti hai; ek card jo gently curved hai (cambered) aur bhi zyada hawa ko neeche scoop karta hai, chahe wo zyada tili na ho. Hum imagine karte hain ki wing tiny spinning straws ki ek row hai jo hawa ko pakad kar neeche twist karti hai. Sab spinning ko count karo, speed se aur hawa ke weight se multiply karo, aur wahi hai tumhari lift. Tilt ( $α$ ) aur curve ( $β$ ) do tarike hain isse aur zyada scoop karwane ke.

Connections

Kutta–Joukowski theorem — $L^{'} = ρ V Γ$ deta hai.
Circulation and bound vortices — $Γ$ ka origin.
Kutta condition — physical $γ (x)$ select karta hai.
Biot–Savart law — induced downwash $w (x)$ deta hai.
Glauert's integral and Fourier coefficients — sheet solve karta hai.
Lift coefficient and angle of attack — $c_{ℓ}$ – $α$ curve.
Compressibility corrections (Prandtl–Glauert) — incompressible se aage ka step.