Angle of attack, lift coefficient, drag coefficient: Question bank

3.1.20 · D5 · Physics › Compressible Flow & Aerodynamics › Angle of attack, lift coefficient, drag coefficient

0. Symbols, axes, aur pictures jo pehle chahiye

Traps se pehle, teen quick self-contained anchors taaki neeche koi bhi symbol mystery na rahe.

Definition Lift-curve ke symbols

Relative wind $V_{\infty}$ : air ki velocity far upstream, yani wo direction jis taraf se air wing ke apne frame mein aati lage. Level flight mein yeh horizontal hoti hai aur flight path ke saath backward point karti hai. Yahi sab cheez ka reference direction hai — neeche figure dekho.
Chord line: airfoil ke leading edge se trailing edge tak seedhi line.
Angle of attack $α$ : relative wind se chord line tak ka angle. Positive $α$ = leading edge hawa mein upar ki taraf jhuka hua.
Lift $L$ : force component jo relative wind ke perpendicular ho. Positive lift airfoil ke upper (suction) side ki taraf point karti hai. Drag $D$ : relative wind ke parallel component, downstream ki taraf point karta hai.
Zero-lift angle $α_{L = 0}$ : woh specific $α$ jis par $C_{L} = 0$ ho. Symmetric section ke liye $α_{L = 0} = 0$ ; normally-cambered wing ke liye $α_{L = 0} < 0$ (woh $α = 0$ par bhi lift deta hai).
Lift-curve slope $a_{0} = \frac{d C _{L}}{d α}$ : $C_{L}$ kitni tezi se per unit angle climb karta hai. Iske units hote hain — ya toh per radian ( $2 π \approx 6.28/ rad$ thin-airfoil theory se) ya per degree ( $\approx 0.11/ deg$ ). Hamesha batao konsa use kar rahe ho; dono $π /180 \approx 1/57.3$ ke factor se differ karte hain.

Figure — Angle of attack, lift coefficient, drag coefficient

True or false — justify karo

Zero angle of attack par ek flat plate zero lift produce karta hai.

Symmetric shape ke liye True. Ek symmetric section (ya flat plate) ka

α_{L = 0} = 0

hota hai, toh

α = 0

par

C_{L} = a_{0} (α - 0) = 0

. Ek cambered airfoil ka

α_{L = 0} < 0

hota hai, toh woh

α = 0

par bhi lift banata hai.

Fixed angle of attack par airspeed double karne se lift double ho jaati hai.

False. Lift

q = \frac{1}{2} ρ V^{2}

ke saath scale karti hai, toh

V

double karne se lift

2^{2} = 4

se multiply hoti hai,

2

se nahi.

C_{L}

khud speed ke saath nahi badalta (jab tak Mach effects na aayein).

Lift coefficient

C_{L}

ke units newtons per square metre hain.

False.

C_{L} = L / (\frac{1}{2} ρ V^{2} S)

hai yani force ÷ (pressure × area) = force ÷ force = dimensionless. Yahi dimensionlessness exactly isliye hai ki ek wind-tunnel model full-scale wing predict kar sakta hai.

Lift hamesha ground ke relative vertically upward act karti hai.

False. Upar diye sign convention ke hisaab se, lift relative wind ke perpendicular hoti hai, gravity ke nahi. Climb, dive, ya bank mein relative wind tilt hoti hai, toh lift bhi saath tilt hoti hai — banked turns isi tarah kaam karte hain.

Stall angle par lift apne maximum par hoti hai.

True.

C_{L}

α

ke saath

α_{stall}

tak badhta hai jahan woh peak (

C_{L, m a x}

) karta hai; us angle ke aage woh drop karta hai. Toh stall angle maximum lift mark karta hai, zero lift nahi — Boundary Layer Separation & Stall dekho.

Induced-drag force ko sirf tez fly karke chota kiya ja sakta hai.

Ek limit ke taur par True, lekin dhyan se padho. Fixed weight

W

hold karne ke liye,

C_{L}

W = C_{L} q S

satisfy karna hoga, toh

C_{L} \propto 1/ V^{2}

; phir induced-drag force

D_{i} = C_{D, i} q S \propto C_{L}^{2} \cdot q \propto (1/ V^{2})^{2} \cdot V^{2} = 1/ V^{2} \to 0

jaise

V \to \infty

. Jo cheez nonzero lift par kabhi zero nahi ho sakti woh hai induced-drag coefficient

C_{D, i} = C_{L}^{2} / (π e AR)

— coefficient aur force ko confuse mat karo (Induced Drag & Wingtip Vortices dekho).

Fixed lift coefficient par aspect ratio badhane se induced drag kam hoti hai.

True. Induced drag

= C_{L}^{2} / (π e AR)

hai, toh bada

AR

(lamba, patla wing) isko kam karta hai — isliye gliders ke wings bahut lambe hote hain.

Profile drag

C_{D, 0}

attack ke sabhi angles ke liye exactly constant hota hai.

False (sirf low $α$ par approximately true).

C_{D, 0}

linear region mein roughly constant hota hai, lekin stall ke paas separated flow wake ko bada kar deta hai aur form drag aur total

C_{D}

dono tezi se badhte hain.

Angle of attack aircraft ke pitch angle ke barabar hota hai.

False. Pitch horizon ke against measure hota hai; angle of attack relative wind ke against. Ek plane nose-level descend karta hua bhi positive

α

rakhta hai kyunki hawa thodi neeche se aati hai.

Error dhundho

"Kyunki force

ρ V^{2} S

hai, hum likh sakte hain

L = C_{L} ρ V^{2} S

Missing

\frac{1}{2}

hi error hai. Lift dynamic pressure

q = \frac{1}{2} ρ V^{2}

use karta hai, toh

L = C_{L} \frac{1}{2} ρ V^{2} S

C_{L}

usi

\frac{1}{2}

ke saath calibrate hota hai; isko drop karne se har coefficient do ke factor se galat ho jaata hai.

α_{L = 0} = - 2^{\circ}

wale wing se

α = 0^{\circ}

par koi lift nahi hoti."

Galat. Per-degree slope

a_{0} \approx 0.11/ deg

use karte hue taaki angles degrees mein rahein,

C_{L} = a_{0} (α - α_{L = 0}) = 0.11 \times (0 - (- 2)) = 0.11 \times 2 \approx 0.22 > 0

. (Agar per-radian

a_{0} = 2 π

insist karo, toh pehle

2^{\circ} = 0.0349

rad convert karo, deta hai

2 π \times 0.0349 \approx 0.22

— same answer.) Zero lift

- 2^{\circ}

par hoti hai, toh

α = 0

par yeh cambered wing already lift kar raha hai.

"Thin-airfoil theory ek lift-curve slope of

2 π

per degree deta hai."

2 π

per radian hai, per degree nahi. Convert karo:

2 π / rad \times (π /180) rad/deg \approx 0.11

per degree. Units mix karne se slope approximately 57 ke factor se inflate ho jaata hai.

"Tez planes ka lift coefficient zyada hota hai."

C_{L}

α

par depend karta hai (aur secondarily Mach/Reynolds par), directly

V

par nahi. Tez flight

q

badhata hai, toh lift badhti hai chahe

C_{L}

fixed

α

par unchanged ho.

"Kutta–Joukowski result

L^{'} = ρ V Γ

kaam karta hai kyunki pressure wing ko upar push karta hai."

Subtle slip:

L^{'} = ρ V Γ

ek statement hai circulation

Γ

ke baare mein jo Kutta-Joukowski Theorem aur Kutta condition se set hota hai, na ki direct pressure balance se. Pressure mechanism hai; circulation woh bookkeeping hai jo total force predict karta hai.

"Kyunki

L / D = C_{L} / C_{D}

, sabse zyada possible

C_{L}

par fly karne se best glide milti hai."

Galat — high

C_{L}

induced drag bhi blow up karta hai (

\propto C_{L}^{2}

), toh

C_{D}

C_{L}

se eventually tez badhta hai. Best

L / D

ek intermediate

C_{L}

par hoti hai jahan profile aur induced drag balanced hain,

C_{L, m a x}

par nahi.

Why questions

Hum force versus speed tabulate karne ki jagah dimensionless coefficient define karne ki koshish kyun karte hain?

Kyunki force har speed, density, aur size ke saath badlti hai, lekin

C_{L} (α)

shape ka DNA hai — ek chhote model par ek baar measure karo aur

L = C_{L} q S

se kisi bhi

V

ρ

, aur

S

par lift predict karo.

Stall angle par lift gradually fade hone ki bajay suddenly kyun marti hai?

Boundary layer attached aur orderly rehti hai jab tak adverse pressure gradient bahut steep na ho jaaye; phir woh ek large region par almost ek saath separate ho jaata hai, upper surface par low-pressure suction collapse kar deta hai — yeh ek threshold effect hai, Boundary Layer Separation & Stall ke hisaab se.

q = \frac{1}{2} ρ V^{2}

mein woh

\frac{1}{2}

wahi kyun hai jo Bernoulli's Equation mein appear karta hai?

Kyunki

q

hi flow ki kinetic-energy density hai, aur Bernoulli ka pressure balance

p + \frac{1}{2} ρ V^{2} = const

exactly woh KE term ko static pressure ke against trade karta hai — aerodynamic coefficients isko directly inherit karte hain.

Lift force banana wing ko induced drag kyun banata hai?

Lift generate karna wingtip vortices shed karta hai jo local flow ko ek induced angle

α_{i} \propto C_{L}

se downward (downwash) tilt karte hain; yeh lift vector ko backward rotate karta hai, aur uska backward component

D_{i} \approx L α_{i} \propto C_{L}^{2}

hi induced drag hai — vertical force ke bina tilt nahi ho sakta (Induced Drag & Wingtip Vortices dekho).

Thin-airfoil theory shape se independent ek universal slope

2 π

(per radian) kyun deta hai?

Slope ek flat vortex sheet ki linearised circulation aur Kutta condition se aata hai, jo first order tak geometry-independent hain. Camber aur thickness

α_{L = 0}

shift karte hain aur

C_{L, m a x}

change karte hain lekin leading-order slope ko

2 π

per radian ke paas chhod dete hain.

Agar coefficients "dimensionless" hain toh bhi Reynolds aur Mach number par depend kyun karte hain?

Dimensionless matlab units se free, sari physics se nahi. Reynolds Number boundary-layer behaviour govern karta hai aur Mach Number compressibility govern karta hai — dono pressure field reshape karte hain, toh

C_{L}

aur

C_{D}

fixed

α

par bhi unke saath shift karte hain.

Ek symmetric airfoil zero angle of attack par kabhi lift kyun nahi kar sakta, jabki cambered wala kar sakta hai?

Symmetry

α = 0

par upper aur lower flow ko mirror images hone par force karti hai, toh pressures cancel ho jaate hain aur

C_{L} = 0

. Camber us mirror ko tod deta hai, net upward suction deta hai chahe chord hawa ke saath aligned ho (

α_{L = 0} < 0

Edge cases

Zero-lift angle

α = α_{L = 0}

par exactly

C_{L}

kya hai?

Exactly zero:

C_{L} = a_{0} (α_{L = 0} - α_{L = 0}) = 0

, chahe

a_{0}

per-radian ho ya per-degree. Yahi poore lift curve ka reference point hai — effective angle yahin se measure hota hai, chord se nahi.

V = 0

par (aircraft at rest) lift aur drag ka kya hoga?

Dono vanish ho jaate hain kyunki

q = \frac{1}{2} ρ (0)^{2} = 0

, toh

L = C_{L} q S = 0

aur similarly

D = 0

. Koi relative wind nahi matlab koi aerodynamic force nahi, angle of attack chahe kuch bhi ho — isliye takeoff ke liye ground run chahiye.

Negative angle of attack par (

α_{L = 0}

se neeche), lift kis direction mein point karti hai?

Sign convention ke hisaab se lift negative ho jaati hai ("belly"/lower surface ki taraf point karti hai), kyunki

C_{L} = a_{0} (α - α_{L = 0}) < 0

jab

α < α_{L = 0}

. Inverted flight aur downforce wings exactly isi sign ko exploit karte hain.

Agar ek wing ka infinite aspect ratio hota, toh induced drag ka kya hota?

Woh zero ho jaata:

C_{L}^{2} / (π e AR) \to 0

jaise

AR \to \infty

. Yeh idealised two-dimensional airfoil hai bina tips ke, hence koi tip vortices nahi — woh limiting case jo thin-airfoil theory assume karta hai.

Deep-stall region mein stall angle ke kafi aage, kya

C_{L}

zero hota hai?

Nahi —

C_{L}

drop hota hai lekin positive rehta hai (ek flat plate high angle par bhi kuch air deflect karta hai); woh bas

C_{L, m a x}

se kafi neeche gir jaata hai jabki

C_{D}

tezi se badhta hai,

L / D

ratio bigad deta hai.

Jaise Mach number 1 ke paas aata hai, constant-slope thin-airfoil result ko kya chahiye?

Ek compressibility correction. Prandtl–Glauert factor slope ko multiply karta hai, toh

a_{0}

ab fixed

2 π

per radian nahi rehta — low-speed formula sirf tabhi hold karta hai jab flow effectively incompressible ho, Compressible Flow ke hisaab se.

Exactly stall angle par, kya flow fully separated hai?

Fully nahi — separation

C_{L}

ke peak par shuru hona shuru ho rahi hoti hai. Full separation stall angle ke baad hoti hai; bilkul peak par, attached aur separated flow ek aisi losing balance mein hain jo jaise

α

aur badhta hai, tilt ho jaati hai.

Connections

Angle of attack, lift coefficient, drag coefficient — woh parent jise yeh bank test karta hai.
Bernoulli's Equation — dynamic pressure ke saath shared $\frac{1}{2}$ .
Kutta-Joukowski Theorem — $2 π α$ slope ke peeche.
Boundary Layer Separation & Stall — stall traps.
Induced Drag & Wingtip Vortices — $C_{L}^{2}$ traps.
Reynolds Number & Mach Number — kyun dimensionless hone ka matlab phir bhi shape-dependent hai.
Compressible Flow — high-Mach edge case.