Aerodynamic coefficients — CN, CA, CL, CD, Cm as functions of angle of attack, Mach: Exercises

3.1.29 · D4 · Physics › Compressible Flow & Aerodynamics › Aerodynamic coefficients — CN, CA, CL, CD, Cm as functions o

Is page par use kiye gaye symbols (sab parent mein define hain): $C_{N}$ normal-force coefficient, $C_{A}$ axial-force coefficient, $C_{L}$ lift coefficient, $C_{D}$ drag coefficient, $C_{m}$ moment coefficient, $α$ angle of attack (chord aur freestream $V_{\infty}$ ke beech ka angle), $q_{\infty}$ dynamic pressure, $S$ reference area, $c$ reference chord, $M_{\infty}$ freestream Mach number, $β = 1 - M_{\infty}^{2}$ (subsonic) ya $M_{\infty}^{2} - 1$ (supersonic).

Level 1 — Recognition

Recall Solution

KYA karte hain: numbers ko definition mein plug karo — arithmetic ke aage koi physics nahi. $q_{\infty} = \frac{1}{2} (1.225) (50)^{2} = \frac{1}{2} (1.225) (2500) = 1531.25 Pa .$ KYUN: $q_{\infty}$ flow ka "pressure scale" hai; har force coefficient isse divide karta hai taaki size aur speed nikal jaayein.

Recall Solution

(a) $M_{\infty} = 0.5$ subsonic hai → Prandtl–Glauert: incompressible $C_{L}$ ko $1 - M_{\infty}^{2} = 1 - 0.25 = 0.75 \approx 0.866$ se divide karo. (b) $M_{\infty} = 2.0$ supersonic hai → Ackeret: $M_{\infty}^{2} - 1 = 4 - 1 = 3 \approx 1.732$ se divide karo. WHY flip hota hai: wahi square root appear hoti hai, lekin $M = 1$ cross karte waqt uske andar ka sign badal jaata hai. $M = 1$ ke neeche woh factor lift ko amplify karta hai ( $< 1$ , toh divide karne se bada ho jaata hai); $M = 1$ ke upar factor $> 1$ hota hai, toh Mach badhne ke saath lift kam hoti hai.

Recall Solution

$q_{\infty} S$ ki units hain $(Pa) (m^{2}) = (N/m^{2}) (m^{2}) = N$ — bilkul ek force. Ek force $L$ ko force se divide karo toh dimensionless milta hai → $C_{L}$ kaam karta hai. Ek moment $M$ ki units hain $N \cdot m$ . $q_{\infty} S$ (units N) se divide karne par metres bachte hain, pure number nahi milta. Humein ek aur length $c$ chahiye, taaki $q_{\infty} S c$ ki units $N \cdot m$ ho jaayein, jo moment se match kare. KYUN important hai: $c$ bhool jaane par ek "coefficient" milta hai jisme abhi bhi units hain aur jo model scale ke saath change ho jaata hai — jo iska poora purpose hi khatam kar deta hai.

Level 2 — Application

Recall Solution

KYA: force components ko body axes se wind axes mein rotate karo (figure dekho — wahi vector, do rulers $α$ se rotate hue).

Figure — Aerodynamic coefficients — CN, CA, CL, CD, Cm as functions of angle of attack, Mach

cos 1 0^{\circ} = 0.9848

sin 1 0^{\circ} = 0.1736

C_{L} = C_{N} cos α - C_{A} sin α = 1.10 (0.9848) - 0.08 (0.1736) = 1.0833 - 0.0139 = 1.069 .

C_{D} = C_{N} sin α + C_{A} cos α = 1.10 (0.1736) + 0.08 (0.9848) = 0.1910 + 0.0788 = 0.270 .

KYUN: normal force

N

, lift direction se

α

thoda tilt hota hai, toh uska zyada hissa (

C_{N} cos α

) lift banta hai jabki ek chhota aft-pointing axial slice (

- C_{A} sin α

) subtract hota hai. Dono forces streamwise direction mein contribute karte hain, isliye drag unhe add karta hai.

Recall Solution

Pehle radians mein convert karo (slope $2 π$ per radian hai): $α = 6^{\circ} \times \frac{π}{180} = 0.10472 rad$ . $C_{L} = 2 π (α - α_{L = 0}) = 2 π (0.10472) = 0.658 .$ KYUN radians: $2 π$ slope Thin Airfoil Theory mein chord par ek integral se aata hai jo naturally radians mein hai. Degrees plug karne par answer ~57× inflate ho jaata.

Recall Solution

$β = 1 - M_{\infty}^{2} = 1 - 0.49 = 0.51 = 0.7141$ . $C_{L} = \frac{C _{L, incomp}}{β} = \frac{0.658}{0.7141} = 0.921 .$ KYUN divide karne se amplify hota hai: jab flow $M = 1$ ki taraf speed karta hai, density variations usi shape ko zyada air deflect karne deti hain, toh lift per degree badhti hai. Dekho Prandtl-Glauert Compressibility Correction.

Recall Solution

$α = 3^{\circ} = 0.05236 rad$ , $M_{\infty}^{2} - 1 = 6.25 - 1 = 5.25 = 2.2913$ . $C_{L} = \frac{4 α}{M _{\infty}^{2} - 1} = \frac{4 ( 0.05236 )}{2.2913} = \frac{0.20944}{2.2913} = 0.0914 .$ $C_{D, wave} = \frac{4 α ^{2}}{M _{\infty}^{2} - 1} = \frac{4 ( 0.05236 ) ^{2}}{2.2913} = \frac{0.010966}{2.2913} = 0.00479 .$ KYUN wave drag exist karta hai: supersonically, plate shock/expansion waves banati hai jo energy le jaati hain — ek aisa drag jiska koi subsonic counterpart nahi, aur yeh $α^{2}$ ke saath scale hota hai. Dekho Supersonic Linearized (Ackeret) Theory.

Level 3 — Analysis

Recall Solution

Induced-drag term: $\frac{C _{L}^{2}}{π e A R} = \frac{0. 9 ^{2}}{π ( 0.85 ) ( 8 )} = \frac{0.81}{21.362} = 0.03792$ . $C_{D} = C_{D 0} + \frac{C _{L}^{2}}{π e A R} = 0.020 + 0.03792 = 0.0579 .$ Induced fraction $= \frac{0.03792}{0.0579} = 0.655$ yani lagbhag $65.5%$ . KYUN parabolic hai: kyunki $C_{L} \propto α$ hai, induced term $\propto C_{L}^{2} \propto α^{2}$ hota hai — drag-vs- $α$ curve upar khulta hua parabola hai. Dekho Induced Drag and Wingtip Vortices.

Recall Solution

Inverse rotation use karo ( $α$ se doosri taraf rotate karo): $cos 1 2^{\circ} = 0.9781$ , $sin 1 2^{\circ} = 0.2079$ . $C_{N} = C_{L} cos α + C_{D} sin α = 0.75 (0.9781) + 0.045 (0.2079) = 0.73359 + 0.00936 = 0.743 .$ $C_{A} = - C_{L} sin α + C_{D} cos α = - 0.75 (0.2079) + 0.045 (0.9781) = - 0.15593 + 0.04401 = - 0.112 .$ WHAT negative $C_{A}$ ka matlab: negative axial coefficient chord ke saath aage point karta hai — yahan lifting force ka strong forward-tilted suction chhoti backward friction drag ko overwhelm karta hai. Moderate $α$ par lifting airfoil ke liye yeh physically normal hai (leading-edge suction).

Recall Solution

$C_{N} sin α = 0.743 (0.2079) = 0.15447$ . $C_{A} cos α = - 0.112 (0.9781) = - 0.10955$ . Sum $= 0.15447 - 0.10955 = 0.0449 \approx 0.045 ✓$ (diye gaye $C_{D}$ se match karta hai). Interpretation: normal-force tilt ek positive streamwise push contribute karta hai (jaise induced drag), jabki forward-pointing axial term subtract karta hai (leading-edge suction thrust recover karta hai). Net drag woh chhota residual hai — exactly isi liye inviscid lifting airfoils bahut low drag rakh sakte hain.

Level 4 — Synthesis

Recall Solution

(a) Effective angle $= α - α_{L = 0} = 5^{\circ} - (- 2^{\circ}) = 7^{\circ} = 0.12217 rad$ . $C_{L, incomp} = 2 π (0.12217) = 0.7676$ . (b) $β = 1 - 0.36 = 0.64 = 0.8$ . $C_{L} = 0.7676/0.8 = 0.9595$ . (c) $C_{N} \approx C_{L} = 0.9595$ , $C_{A} = 0.010$ , $α = 5^{\circ}$ ( $sin = 0.08716$ , $cos = 0.99619$ ) ke saath: $C_{D} = C_{N} sin α + C_{A} cos α = 0.9595 (0.08716) + 0.010 (0.99619) = 0.08363 + 0.00996 = 0.0936 .$ KYUN approximation theek hai: $5^{\circ}$ par, $cos α \approx 1$ toh $C_{L} \approx C_{N}$ $0.4%$ se kam error ke saath — parent note se small-angle sanity check.

Recall Solution

(a) $β = 1 - 0.9025 = 0.0975 = 0.31225$ . $C_{L} = 0.5/0.31225 = 1.601$ . (b) Formula lift ko $1/ β \approx 3.2 \times$ amplify karta hai — ek bahut bada jump. Lekin $M_{\infty} = 1$ ke paas, airfoil ke upar local flow already supersonic ho jaata hai, shocks form ho jaate hain, aur linearized subsonic potential-flow assumption toot jaati hai. Jab $M_{\infty} \to 1$ , $β \to 0$ aur formula infinite lift predict karta hai, jo physically impossible hai. Sirf $M_{\infty} ≲ 0.7$ tak valid hai. Lesson: ek formula arithmetically bilkul theek ho sakta hai aur apne domain ke bahar physically nonsense bhi.

Recall Solution

(a) Stability ke liye $\frac{d C _{m}}{d α} < 0$ chahiye. Yahan $\frac{d C _{m}}{d α} = - 0.9 < 0$ → statically stable ✔. KYUN: agar ek gust $α$ badhata hai, toh moment negative ho jaata hai (nose-down), $α$ ko wapas push karta hai — ek restoring response. Dekho Static Longitudinal Stability. (b) Trim jahan $C_{m} = 0$ : $0.06 - 0.9 α = 0 \Rightarrow α_{trim} = 0.06/0.9 = 0.06 \overline{6} rad = 3.8 2^{\circ}$ . KYUN important hai: $C_{m 0} > 0$ (positive intercept) plus negative slope ek positive, flyable trim angle guarantee karta hai — aircraft naturally $3.8 2^{\circ}$ par settle ho jaata hai.

Level 5 — Mastery

Recall Solution

KYA: AC woh point hai jahan $C_{m}$ $C_{L}$ ke saath nahi badalta (yani $α$ ke saath bhi nahi). $C_{m, ac}$ hatane ke liye do equations subtract karo: $C_{m, LE} (0.8) - C_{m, LE} (0.4) = - \frac{x _{a c}}{c} (0.8 - 0.4) .$ $- 0.240 - (- 0.140) = - 0.100 = - \frac{x _{a c}}{c} (0.4) \Rightarrow \frac{x _{a c}}{c} = \frac{0.100}{0.4} = 0.25 .$ $C_{L} = 0.4$ par back-substitute karo: $- 0.140 = C_{m, ac} - 0.25 (0.4) = C_{m, ac} - 0.100$ $\Rightarrow C_{m, ac} = - 0.040$ . KYUN $0.25$ famous hai: thin-airfoil theory predict karta hai ki AC quarter-chord par hota hai — yeh data isse confirm karta hai. Dekho Aerodynamic Center vs Center of Pressure.

Recall Solution

KYUN yahan derivative: $L / D$ ek ratio hai jo $C_{L}$ ke saath pehle badhta hai phir girta hai; uska peak wahan hai jahan slope zero ho. $f (C_{L}) = \frac{C _{L}}{C _{D 0} + k C _{L}^{2}}$ maximize karo. $f^{'} = 0$ set karo:

\Rightarrow C_L^\star = \sqrt{\frac{C_{D0}}{k}}.$$ $$C_L^\star = \sqrt{\frac{0.020}{0.0468}} = \sqrt{0.4274} = \mathbf{0.6538}.$$ Is point par $kC_L^{\star 2} = C_{D0}$ (induced drag parasite drag ke **barabar** hai — ek key result): $C_D = 0.020 + 0.020 = 0.040$, toh $$\left(\frac{L}{D}\right)_{\max} = \frac{0.6538}{0.040} = \mathbf{16.35}.$$ **KYUN yeh beautiful hai:** best glide exactly tab hota hai jab do drag sources balanced hote hain.

Recall Solution

(a) Thin-airfoil: $\frac{d C _{L}}{d α} = 2 π = 6.283$ /rad. (b) Prandtl–Glauert slope ko $1/ β$ se scale karta hai: $β = 1 - 0.64 = 0.6$ , $\frac{d C _{L}}{d α} = 2 π /0.6 = 10.47$ /rad. (c) Ackeret flat plate: $C_{L} = 4 α / M^{2} - 1$ , toh $\frac{d C _{L}}{d α} = \frac{4}{3} = 2.309$ /rad.

Trend (KYUN hump hai): subsonically slope

M = 1

ki taraf badhta hai (compressibility har degree par zyada lift pack karta hai). Supersonically slope collapse ho jaata hai —

M^{2} - 1

denominator badhta hai, toh wahi shape far less lift per degree banati hai. Dono ke beech ek discontinuous "transonic gap" hai jise koi bhi linear theory cover nahi karta.

Recall Har formula ki ek-line summary

$q_{\infty} = \frac{1}{2} ρ V^{2}$ · $C_{L} = C_{N} cos α - C_{A} sin α$ · $C_{D} = C_{N} sin α + C_{A} cos α$ · $C_{L} = 2 π (α - α_{L = 0})$ · PG: $/ 1 - M^{2}$ · Ackeret: $C_{L} = 4 α / M^{2} - 1$ , $C_{D, w} = 4 α^{2} / M^{2} - 1$ · polar $C_{D} = C_{D 0} + k C_{L}^{2}$ · AC jahan $d C_{m} / d C_{L} = 0$ · best $L / D$ jahan $k C_{L}^{2} = C_{D 0}$ .