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

3.1.21 · D5 · Physics › Compressible Flow & Aerodynamics › Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c

Yahan reuse hone wale symbols (sab parent mein banaye gaye hain): $L^{'}$ = lift per unit span (N/m), $Γ$ = circulation, $α$ = angle of attack radians mein, $β$ = camber slope parameter, $c$ = chord, $V$ = freestream speed, $ρ$ = density, $c_{ℓ}$ = section lift coefficient, $γ (x)$ = vortex-sheet strength, $A_{0}, A_{1}, \dots$ = Glauert Fourier coefficients.

True or false — justify

Camber lift-curve ko steeper banata hai (bada

d c_{ℓ} / d α

False. Slope

2 π

per radian par fix rehta hai; camber

c_{ℓ} = 2 π α + 4 β

mein sirf constant

4 β

add karta hai, poori line ko left slide karta hai, kabhi tilt nahi karta.

Ek symmetric (uncambered) airfoil

α = 0

par zero lift produce karta hai.

True.

d z / d x = 0

ke saath hume

A_{0} = α

A_{1} = 0

milta hai, isliye

c_{ℓ} = 2 π α

, jo

α = 0

par

0

hai — camber nahi matlab bina tilt ke lift nahi.

Ek cambered airfoil

α = 0

par zero lift produce karta hai.

False. Uska zero-lift angle

α_{L = 0} = - 2 β / π < 0

hai, isliye

α = 0

par bhi lift rehti hai; lift khatam karne ke liye use thoda neeche tilt karna padta hai.

Freestream speed

V

double karne se lift per unit span double ho jaati hai.

False.

L^{'} = ρ V Γ

hai lekin

Γ

khud

V

ke proportional hai (dekho

Γ = π c V (A_{0} + A_{1} /2)

), isliye

L^{'} \propto V^{2}

—

V

double karne se

L^{'}

chaar guna ho jaati hai.

Lift airfoil ki poori curved surface par depend karti hai, isliye hum use ek straight line of vortices se replace nahi kar sakte.

False. Thin airfoil theory vortex sheet ko mean camber line par rakhti hai (thin wing ke liye almost straight); yeh same outer flow reproduce karta hai kyunki airfoil stream ko bahut kam disturb karta hai.

Saare Fourier coefficients

A_{0}, A_{1}, A_{2}, A_{3}, \dots

lift mein contribute karte hain.

False. Sirf

A_{0}

aur

A_{1}

circulation integral mein bachte hain;

n \geq 2

ke liye

sin n θ

terms orthogonality ki wajah se zero integrate hote hain, isliye higher coefficients moment ko affect karte hain, lift ko nahi.

Kutta–Joukowski

L^{'} = ρ V Γ

ek approximation hai jo sirf thin airfoils ke liye valid hai.

False. Kutta–Joukowski circulation

Γ

wale kisi bhi 2-D body ke liye inviscid incompressible flow mein exact hai; thin-airfoil theory sirf woh approximate tarika hai jisse hum

Γ

estimate karte hain.

Agar circulation

Γ

zero hai, toh plate ke tilt se section phir bhi kuch lift produce karta hai.

False.

L^{'} = ρ V Γ

hai, isliye

Γ = 0

se exactly zero lift milti hai chahe

α

kuch bhi ho — no circulation, no lift, bas itna hi.

Kutta condition ek extra physical assumption hai jo hum impose karte hain, equations khud force nahi karti.

True. Inviscid theory akele infinitely many circulations allow karti hai; Kutta condition (sharp trailing edge par finite velocity) woh single physical wala select karti hai, jo viscosity actually kya karti hai uski nakal karte hue.

Spot the error

"Kyunki hum wing ko

α = 5^{\circ}

tilt karte hain, hum

α = 5

c_{ℓ} = 2 π α

mein plug karte hain."

Error: units. Linearisation mein

sin α \approx α

radians mein use hua tha; pehle

5^{\circ} = 0.0873

rad convert karna zaroori hai, warna lift

\approx 57

ke factor se inflate ho jaati hai.

"Camber lift mein

\frac{2 β}{π}

contribute karta hai, aur kyunki

β

ek angle hai isliye hume ise

α

ki tarah degrees mein express karna chahiye."

Error: consistency.

β

ek slope/angle parameter hai jo same linearised algebra mein enter karta hai, isliye yeh bhi radian-equivalent quantity hona chahiye; ek ke liye degrees aur doosre ke liye radians mix karna sum

α + 2 β / π

ko corrupt kar deta hai.

"Top wali air zyada lambi path travel karti hai aur trailing edge par bottom wali air se milni chahiye, isliye yeh speed up hoti hai — yahi lift hai."

Error: the equal-transit-time myth. Top aur bottom ke particles saath nahi aate; asli wajah bound circulation hai jo Kutta condition set karti hai, jo genuinely upper flow ko faster banati hai.

"Vortex sheet strength

γ (θ) = 2 V (A_{0} \frac{1 + c o s θ}{s i n θ} + \sum A_{n} sin n θ)

hai, aur trailing edge par yeh blow up ho jaata hai."

Error. Trailing edge par

θ = π

par,

\frac{1 + c o s θ}{s i n θ} \to 0

(numerator aur

sin θ

dono

\to 0

, ratio

\to 0

), isliye

γ

wahan finite rehta hai — yahi form Kutta condition ko encode karta hai.

"Kyunki

L^{'} = ρ V Γ

mein

ρ

hai, vacuum mein wing apni spin se thodi lift produce karti rahegi."

Error. Vacuum mein

ρ = 0

L^{'} = 0

milta hai; momentum neeche carry karne ke liye koi fluid nahi hai, isliye chahe "sheet" kuch bhi spin kare, lift zero hi rehegi.

L^{'}

poori wing ki total lift hai."

Error: the prime.

L^{'}

lift per unit span (N/m) hai. Total lift ke liye

L^{'} \times

span chahiye (saath mein finite-wing 3-D effects ka correction, jo yeh 2-D theory ignore karti hai).

Γ

find karne ke liye hum downwash

w (x)

ko chord ke along integrate karte hain."

Error.

Γ = \int_{0}^{c} γ d x

mein vortex-sheet strength

γ

integrate hoti hai, downwash

w

nahi;

w

sirf flow-tangency condition mein appear hota hai jo

γ

find karne ke liye use hoti hai.

Why questions

Why does the derivation put the vortices on the camber line instead of the actual airfoil surface?

Thin airfoil flow ko bahut kam perturb karta hai, isliye first order par mean line aur surface outer stream ke liye indistinguishable hote hain; line use karne se integral 1-D ho jaata hai aur analytically solvable ban jaata hai.

Why do only

A_{0}

and

A_{1}

appear in the lift, while

A_{2}, A_{3}, \dots

vanish?

Circulation integral

\int_{0}^{π} γ d θ

mein

[0, π]

par

sin n θ

ki orthogonality aati hai;

n \geq 2

ka har term zero integrate hota hai, sirf

A_{0}

aur

A_{1}

contributions bachti hain.

Why must the angle of attack be small for this theory?

Humne

sin α \approx α

linearise kiya aur sheet ko flat treat kiya; bade angles dono approximations tod dete hain aur physically flow separation (stall) cause karte hain jo inviscid theory capture nahi kar sakti.

Why is the lift-curve slope universally

2 π

regardless of camber shape?

Slope sirf

A_{0} = α - (camber integral)

term ki

α

-dependence se aata hai; camber

A_{0}

aur

A_{1}

ke constant part ko change karta hai lekin yeh

α

pe kaise respond karte hain use nahi, isliye

d c_{ℓ} / d α = 2 π

hamesha rehta hai.

Why does camber let a wing lift at zero angle of attack?

Curvature bina tilt ke bhi air ko neeche deflect karti hai,

c_{ℓ}

mein constant

4 β

add karke; aircraft phir level cruise kar sakta hai aur phir bhi upward force generate karta hai.

Why do we need the Kutta condition at all — isn't the flow uniquely determined?

Pure inviscid theory circulation mein different solutions ka poora family allow karti hai; Kutta condition (sharp edge par smooth departure) woh wala pick karti hai jo real viscosity enforce karti,

Γ

ko uniquely fix karke.

Why does the Biot–Savart law enter the derivation?

Hume yeh jaanna hai ki har vortex element doosre chord points par kitna downwash induce karta hai flow-tangency enforce karne ke liye; Biot–Savart (yahan 2-D

v = Γ/2 π r

) exactly woh tool hai jo vorticity distribution se induced velocity deta hai.

Why is

c_{ℓ}

dimensionless while

L^{'}

has units?

c_{ℓ} = L^{'} / (\frac{1}{2} ρ V^{2} c)

lift ko dynamic pressure times chord se divide karta hai, saare units cancel ho jaate hain isliye same

c_{ℓ}

geometrically similar wings ko kisi bhi scale ya speed par describe karta hai.

Edge cases

What is

L^{'}

for a symmetric flat plate at

α = 0

Zero:

A_{0} = 0

A_{1} = 0

, isliye

Γ = 0

aur

L^{'} = ρ V \cdot 0 = 0

— flow ke saath aligned flat plate koi lift nahi banata.

What is the lift at the zero-lift angle

α = - 2 β / π

for a cambered wing?

Exactly zero — woh angle defined hi aise hai ki

α

term camber term ko cancel kar de, isliye

c_{ℓ} = 0

chahe wing curved ho.

What happens to the formula at

ρ = 0

(vacuum)?

L^{'} = π ρ V^{2} c (\dots) = 0

; koi air nahi toh push karne ke liye kuch nahi, isliye lift vanish ho jaati hai — formula correctly degenerate ho jaata hai.

What happens as the chord

c \to 0

L^{'}

linearly

c

mein

\to 0

jaata hai; infinitely thin-chord "wing" ke paas air turn karne ke liye koi surface nahi, isliye per span koi lift produce nahi hoti.

If we make

α

negative but

β

large enough that

α + 2 β / π > 0

, does the wing still lift up?

Haan — sum sign determine karta hai

Γ

ka; strongly cambered wing moderate negative geometric angle of attack par bhi upward lift kar sakta hai.

At the trailing edge (

θ = π

) what is the vortex-sheet strength

γ

Yeh finite hai (actually zero ki taraf jaati hai); Glauert form precisely isliye choose kiya gaya hai taaki

γ

wahan diverge na kare, Kutta condition enforce karte hue.

What does the theory predict just before stall (large

α

Yeh linear

c_{ℓ} = 2 π α + 4 β

predict karta rehta hai bina kisi upper limit ke — yeh ek known failure hai, kyunki real flow separate ho jaata hai aur lift drop ho jaati hai; inviscid model ke paas stall ka koi mechanism nahi.

If two airfoils have the same

α + 2 β / π

but different camber shapes, do they have the same lift?

Haan, same

L^{'}

, kyunki lift sirf

A_{0} + A_{1} /2

par depend karti hai; different shapes pitching moment mein differ karti hain (higher

A_{n}

se), lift mein nahi.

Recall Traps ka one-line summary

Slope hamesha $2 π$ hai (camber sirf shift karta hai); hamesha radians; lift $ρ V Γ$ hai isliye $\propto V^{2}$ hai aur $ρ$ ya $Γ$ ke saath vanish ho jaati hai; sirf $A_{0}, A_{1}$ lift karte hain; Kutta condition ek imposed physical choice hai, equation ka output nahi.

Connections

Parent: lift per unit span — woh results jinhe ye traps probe karte hain.
Kutta–Joukowski theorem — kyun $L^{'} = ρ V Γ$ exact hai, approximate nahi.
Kutta condition — woh imposed choice jis par kai traps hinge hain.
Circulation and bound vortices — kyun $Γ = 0$ ka matlab zero lift hai.
Biot–Savart law — downwash tool.
Glauert's integral and Fourier coefficients — kyun sirf $A_{0}, A_{1}$ bachte hain.
Lift coefficient and angle of attack — fixed $2 π$ slope.
Compressibility corrections (Prandtl–Glauert) — kya toot jaata hai jab $V$ small nahi hota.