Finite wing theory — induced drag, Prandtl's lifting line: Worked examples

3.1.22 · D3 · Physics › Compressible Flow & Aerodynamics › Finite wing theory — induced drag, Prandtl's lifting line

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Scenario matrix

Is topic ka har problem in cells mein se ek (ya inka blend) hota hai. Har row ek "case class" hai; aakhri column us worked example ka naam hai jo use nail karta hai.

#	Case class	Kya tricky hai	Example
1	Ideal / elliptic ( $δ = 0$ )	Clean formula $C_{D, i} = C_{L}^{2} / (π A R)$	A
2	Non-elliptic ( $δ > 0$ , extra $A_{n}$ )	$δ$ ko harmonics se banana padta hai	B
3	Limiting: $A R \to \infty$	2D infinite wing recover karo	C
4	Degenerate: $C_{L} = 0$	Kya drag vanish hota hai? $C_{L}^{2}$ ka sign	C
5	Lift-slope / effective-AoA	$a = a_{0} / (1 + a_{0} / π A R)$ ki machinery	D
6	Real-world word problem	Weight, speed, area se $C_{L}$ nikalo	E
7	Efficiency factor $e < 1$	$δ \leftrightarrow e$ convert karo, $%$ quote karo	F
8	Exam twist: fixed area, span vary karo	Area constant par $A R$ change hota hai — effect ka sign	G
9	Downwash geometry / sign	Air kahan upar vs neeche jaata hai; $α_{i}$ ka sign	H (figure)

Ab hum cell by cell chalte hain. Steps padhne se pehle guess karo — wahi par learning hoti hai.

Example A — Cell 1: ideal elliptic wing

Worked example Statement

Ek elliptically-loaded wing ka $A R = 8$ hai aur woh $C_{L} = 0.5$ par fly kar raha hai. Iska induced drag coefficient nikalo.

Forecast: Kya $C_{D, i}$ , $C_{L}$ se bada hoga ya chhota? Abhi apne dimaag mein ek rough number guess karo.

Step 1 — formula chuno. Elliptic loading ka matlab sirf $A_{1} \neq = 0$ hai, isliye $δ = \sum_{n \geq 2} n (A_{n} / A_{1})^{2} = 0$ . $C_{D, i} = \frac{C _{L}^{2}}{π A R} .$ Yeh step kyun? $δ = 0$ general $\frac{C _{L}^{2}}{π A R} (1 + δ)$ ko uski sabse clean form mein le aata hai — yeh floor hai, us lift aur span ke liye physically possible sabse kam drag.

Step 2 — numbers daalo. $C_{D, i} = \frac{0. 5 ^{2}}{π \cdot 8} = \frac{0.25}{25.133} = 0.00995.$ Yeh step kyun? Bas arithmetic — lekin notice karo yeh $C_{L} = 0.5$ ke comparison mein kitna tiny hai. Ek lambi wing par lift banana sasta hota hai.

Verify: Units — sab kuch dimensionless hai, achha. Sanity — $C_{D, i} ≪ C_{L}$ , bilkul wahi jo hum ek efficient wing se expect karte hain (lift-to-induced-drag ratio $\approx 50$ ).

Example B — Cell 2: non-elliptic wing

Worked example Statement

Ek wing par $A_{2} / A_{1} = 0.08$ aur $A_{3} / A_{1} = 0.04$ hai (baaki sab $A_{n} = 0$ ), $A R = 6$ aur $C_{L} = 0.6$ ke saath. $C_{D, i}$ nikalo.

Forecast: Kya yeh same $A R$ aur $C_{L}$ wali elliptic wing se better perform karega ya worse? Roughly kitne $%$ se?

Step 1 — $δ$ banao. $δ = \sum_{n \geq 2} n (\frac{A _{n}}{A _{1}})^{2} = 2 (0.08)^{2} + 3 (0.04)^{2} = 0.0128 + 0.0048 = 0.0176.$ Yeh step kyun? $δ$ literally "non-lifting harmonics kitna cost karte hain" yeh batata hai. Har $A_{n}$ ( $n \geq 2$ ) drag add karta hai lekin koi lift nahi (orthogonality unhe lift integral mein khatam kar deti hai).

Step 2 — drag nikalo. $C_{D, i} = \frac{C _{L}^{2}}{π A R} (1 + δ) = \frac{0.36}{π \cdot 6} (1.0176) = 0.01910 \cdot 1.0176 = 0.01944.$ Yeh step kyun? $(1 + δ)$ multiplier elliptic floor ko real value mein convert karta hai.

Verify: Elliptic floor hota $0.36/ (π \cdot 6) = 0.01910$ . Hamara answer $0.01944$ hai, yani $1.76%$ zyada — exactly $δ$ se match karta hai. Consistent. ✓

Example C — Cells 3 & 4: limiting aur degenerate cases

Worked example Statement

(i) Fixed $C_{L}$ par jab $A R \to \infty$ hota hai toh $C_{D, i}$ ka kya hota hai? (ii) Jab $C_{L} = 0$ ho (ek symmetric wing zero lift par) toh $C_{D, i}$ kya hai?

Forecast: Inme se ek exactly zero deta hai. Dono? Guess karo kaun sa.

Step 1 — $A R \to \infty$ limit. $lim_{A R \to \infty} \frac{C _{L}^{2}}{π A R} = 0.$ Yeh step kyun? Infinitely lamba wing ke tips "infinity par" hain — tip ke around leaking effect kuch bhi nahi reh jaata. Hum 2D infinite wing recover karte hain bina induced drag ke, exactly d'Alembert's paradox. Finite theory infinite wali ko limit ke roop mein contain karti hai — poore framework ke liye ek achha sanity check.

Step 2 — $C_{L} = 0$ case. $C_{D, i} = \frac{0 ^{2}}{π A R} = 0.$ Yeh step kyun? Induced drag drag-due-to-lift hai. Lift nahi ⇒ downwash nahi ⇒ rearward tilt nahi ⇒ induced drag nahi. Square note karo: $C_{L}^{2}$ ka matlab $C_{L}$ ka sign kabhi matter nahi karta — neeche ki taraf lift karne wali wing ( $C_{L} < 0$ , jaise inverted race-car wing) wahi induced-drag price pay karti hai jitna upar ki taraf lift karne wali.

Verify: $C_{L} = - 0.5$ , $A R = 8$ ke liye: $C_{D, i} = (- 0.5)^{2} / (π \cdot 8) = 0.00995$ — Example A jitna hi. Down-force ka cost up-force jitna hi hota hai. ✓

Example D — Cell 5: finite-wing lift slope

Worked example Statement

Ek elliptic wing ka $A R = 10$ hai aur 2D lift slope $a_{0} = 2 π$ hai. (i) Iska 3D lift slope $a$ kya hai? (ii) Agar tum $C_{L} = 0.7$ chahte ho aur zero-lift angle $α_{L = 0} = - 2^{\circ}$ hai, toh geometric angle of attack kya set karoge?

Forecast: Kya 3D slope $2 π = 6.283$ se upar hoga ya neeche? Aur kya required angle 2D wing se zyada hoga ya kam?

Step 1 — 3D slope. $a = \frac{a _{0}}{1 + \frac{a _{0}}{π A R}} = \frac{2 π}{1 + \frac{2 π}{10 π}} = \frac{2 π}{1 + 0.2} = \frac{6.2832}{1.2} = 5.236 /rad .$ Yeh step kyun? Geometric angle ka kuch hissa induced angle $α_{i}$ "kha jaata hai", isliye $α$ ka har degree kam lift khareedta hai — slope $2 π$ se neeche gir jaata hai. Yahi finite-wing softening hai.

Step 2 — required angle. $C_{L} = a (α - α_{L = 0})$ se, $α - α_{L = 0} = \frac{C _{L}}{a} = \frac{0.7}{5.236} = 0.13369 rad = 7.66 0^{\circ} .$ Toh $α = α_{L = 0} + 7.66 0^{\circ} = - 2^{\circ} + 7.66 0^{\circ} = 5.66 0^{\circ}$ . Yeh step kyun? Hum lift-slope relation ko invert karte hain "jo lift chahiye" se "jo angle set karna hai" tak jaane ke liye. $- 2^{\circ}$ shift airfoil ki built-in camber account karta hai.

Verify: 2D wing ( $a_{0} = 2 π$ ) ko sirf $0.7/6.2832 = 0.1114$ rad $= 6.38 3^{\circ}$ above $α_{L = 0}$ chahiye hota, yani $α = 4.38 3^{\circ}$ . Finite wing ko zyada angle chahiye ( $5.66 0^{\circ} > 4.38 3^{\circ}$ ) — exactly "softer, needs more incidence" wali kahaani. ✓

Example E — Cell 6: real-world word problem

Worked example Statement

Ek chhote aircraft ka weight $W = 12, 000 N$ hai, wing area $S = 16 m^{2}$ , span $b = 12 m$ , aur woh $ρ = 1.225 kg/m^{3}$ density wali hawa mein $V_{\infty} = 50 m/s$ par cruise kar raha hai. Elliptic loading assume karo. Nikalo (i) $C_{L}$ , (ii) $A R$ , (iii) induced drag force $D_{i}$ newtons mein.

Forecast: Kya $D_{i}$ tens, hundreds, ya thousands of newtons hoga?

Step 1 — force balance se $C_{L}$ . Steady level flight mein, lift = weight: $C_{L} = \frac{W}{\frac{1}{2} ρ V _{\infty}^{2} S} = \frac{12000}{0.5 \cdot 1.225 \cdot 5 0 ^{2} \cdot 16} = \frac{12000}{24500} = 0.4898.$ Yeh step kyun? Dimensionless $C_{L}$ is tarah define hai ki $L = \frac{1}{2} ρ V^{2} S C_{L}$ ; hum use known lift (=weight) se ulta chalate hain.

Step 2 — aspect ratio. $A R = \frac{b ^{2}}{S} = \frac{1 2 ^{2}}{16} = \frac{144}{16} = 9.$ Yeh step kyun? Seedha definition se — span squared over area.

Step 3 — induced drag coefficient, phir force. $C_{D, i} = \frac{C _{L}^{2}}{π A R} = \frac{0.489 8 ^{2}}{π \cdot 9} = \frac{0.23994}{28.274} = 0.008486.$ $D_{i} = \frac{1}{2} ρ V_{\infty}^{2} S C_{D, i} = 0.5 \cdot 1.225 \cdot 2500 \cdot 16 \cdot 0.008486 = 207.9 N .$ Yeh step kyun? Hum dimensionless drag ko usi dynamic-pressure factor $\frac{1}{2} ρ V^{2} S$ se real force mein convert karte hain.

Verify: $\frac{1}{2} ρ V^{2} S$ ke units: $kg/m^{3} \cdot (m/s)^{2} \cdot m^{2} = kg\cdotm/s^{2} = N$ ✓. Sanity: $D_{i} \approx 208 N$ , $12, 000 N$ weight ka $2%$ se kam — sirf induced drag ke liye plausible. ✓

Example F — Cell 7: efficiency factor round-trip

Worked example Statement

Ek wing ka measured Oswald efficiency $e = 0.90$ hai $A R = 7$ , $C_{L} = 0.55$ par. (i) Yeh kis $δ$ ka implication karta hai? (ii) $C_{D, i}$ kya hai? (iii) Ek perfect elliptic wing ke comparison mein, kitna drag $%$ mein waste ho raha hai?

Forecast: $e$ mein $10%$ inefficiency — kya iska matlab exactly $10%$ extra drag hai?

Step 1 — $e$ se $δ$ . Kyunki $e = \frac{1}{1 + δ}$ , $δ = \frac{1}{e} - 1 = \frac{1}{0.90} - 1 = 1.1111 - 1 = 0.1111.$ Yeh step kyun? $e$ aur $δ$ same idea ki do languages hain; $e$ woh hai jo wind-tunnel wale quote karte hain, $δ$ woh hai jo Fourier series produce karta hai.

Step 2 — drag. $C_{D, i} = \frac{C _{L}^{2}}{π e A R} = \frac{0.5 5 ^{2}}{π \cdot 0.90 \cdot 7} = \frac{0.3025}{19.792} = 0.015285.$ Yeh step kyun? Denominator mein $e$ penalty carry karne ka compact tarika hai.

Step 3 — percentage waste. Elliptic floor hai $C_{L}^{2} / (π A R) = 0.3025/ (π \cdot 7) = 0.013756$ . Extra fraction: $\frac{0.015285 - 0.013756}{0.013756} = 0.1111 = 11.11%.$ Yeh step kyun? Yeh exactly $δ$ hai phir se — confirm karta hai ki $δ$ hi fractional drag penalty hai.

Verify: $e = 0.90$ se $11.11%$ extra milta hai, $10%$ nahi — kyunki penalty $1/ e - 1$ hai, $1 - e$ nahi. Yeh ek common trap hai. ✓

Example G — Cell 8: exam twist (fixed area, span badhao)

Worked example Statement

Ek wing ka area $S = 20 m^{2}$ fixed hai, aur tum span ko $b = 10 m$ se $b = 14 m$ tak stretch karte ho (chord mein patla karte hue). Fixed $C_{L} = 0.6$ aur elliptic loading par, $C_{D, i}$ ka kya hoga?

Forecast: Area change nahi hua — toh kya induced drag same rahega, badhega, ya ghategga?

Step 1 — $C_{D, i}$ ko span form mein likho. $C_{D, i} = \frac{C _{L}^{2}}{π A R} = \frac{C _{L}^{2}}{π} \cdot \frac{S}{b ^{2}} .$ Yeh step kyun? $A R = b^{2} / S$ substitute karne se true dependence saamne aati hai: span squared denominator mein, area numerator mein. Span, area nahi, king hai.

Step 2 — dono cases. $A R_{1} = \frac{1 0 ^{2}}{20} = 5, C_{D, i, 1} = \frac{0.36}{π \cdot 5} = 0.022918.$ $A R_{2} = \frac{1 4 ^{2}}{20} = 9.8, C_{D, i, 2} = \frac{0.36}{π \cdot 9.8} = 0.011693.$ Yeh step kyun? Same $S$ , lekin lamba span $A R$ ko $5$ se $9.8$ tak raise karta hai, toh drag girta hai.

Step 3 — ratio. $\frac{C _{D, i, 2}}{C _{D, i, 1}} = \frac{5}{9.8} = 0.5102,$ lagbhag aadha girna. Yeh step kyun? Yeh parent note ke "long & lean" mantra ko confirm karta hai: fixed area par, sirf span badhane se induced drag bahut kum ho jaata hai.

Verify: Area sach mein unchanged hai (dono $20 m^{2}$ ), phir bhi drag aadha ho gaya — exam-trap answer "area fixed ⇒ drag fixed" galat hai. Controlling quantity $b^{2}$ hai. ✓

Example H — Cell 9: downwash geometry aur uska sign (figure ke saath)

Worked example Statement

Ek elliptic wing $C_{L} = 0.6$ produce karta hai $A R = 8$ aur $V_{\infty} = 60 m/s$ par. Elliptic wing ke liye downwash $w$ span ke across uniform hota hai. Nikalo (i) induced angle $α_{i}$ degrees mein, (ii) geometrically confirm karo ki lift rearward tilt hoti hai, aur (iii) $w$ ka sign batao aur kahan flow upar jaata hai.

Forecast: Kya $α_{i}$ degree ka fraction hoga, ya kaafi degrees?

Step 1 — $C_{L}$ aur $A R$ se induced angle. Elliptic wing ke liye $α_{i} = \frac{C _{L}}{π A R}$ (uniform). $α_{i} = \frac{0.6}{π \cdot 8} = 0.023873 rad = 1.36 8^{\circ} .$ Yeh step kyun? Elliptic loading ke liye downwash constant hai, isliye $α_{i} = w / V_{\infty}$ poori wing ke liye ek number hai. Yeh $C_{L} / (π A R)$ ke barabar hai — wahi quantity jo lift-slope formula mein appear hoti hai, jo koi coincidence nahi hai.

Step 2 — geometry padho (figure dekho). Wing $V_{\infty}$ (horizontal) plus neeche ki taraf $w$ se milti hai. Inका sum, local relative wind, $α_{i}$ se neeche point karta hai. Lift us local wind ke $⊥$ hoti hai, isliye lift khud back ki taraf $α_{i}$ se true-vertical se rotate hoti hai. Iska horizontal component induced drag hai: $C_{D, i} = C_{L} α_{i}$ . Yeh step kyun? Yahi induced drag ka physical origin hai — geometry, friction nahi. Figure mein red arrow rearward tilt hai.

Step 3 — us geometric relation ka check. $C_{L} α_{i} = 0.6 \cdot 0.023873 = 0.014324,$ jo $C_{D, i} = \frac{C _{L}^{2}}{π A R} = \frac{0.36}{π \cdot 8} = 0.014324$ ke barabar hona chahiye. Identical. ✓ Yeh step kyun? Yeh prove karta hai ki $C_{D, i} = C_{L} α_{i}$ alag rule nahi hai — yeh usi cheez ko tilted-lift triangle se dekha jaaye toh wahi hai.

Step 4 — sign aur kahan air upar jaati hai. Wing ke peeche aur span ke along, $w$ neeche point karta hai (negative, downwash) — yahi sign humne use kiya. Lekin tips ke baahir, trailing tip vortex doosri taraf circulate karta hai, isliye har wingtip se thoda beyond hawa upar push hoti hai (upwash). Isliye birds V mein urte hain: peeche wala bird apne neighbour ke tip-upwash mein baitha hota hai aur energy bachata hai. Yeh step kyun? Poora sign picture cover karta hai — span ke andar downwash, bahar upwash — jise parent note ne sirf hint kiya tha.

Verify: $α_{i} = 1.3 7^{\circ}$ ek small angle hai, isliye $tan α_{i} \approx α_{i}$ safe hai, aur $C_{D, i} = C_{L} α_{i}$ closed form se last digit tak match kiya. ✓

Recap

Recall Kaun sa formula kaun se cell ke liye?

Elliptic ideal ::: $C_{D, i} = C_{L}^{2} / (π A R)$ with $δ = 0$ Non-elliptic ::: $(1 + δ)$ se multiply karo, $δ = \sum_{n \geq 2} n (A_{n} / A_{1})^{2}$ $A R \to \infty$ ya $C_{L} = 0$ ::: $C_{D, i} \to 0$ 3D lift slope ::: $a = a_{0} / (1 + a_{0} / π A R)$ Coefficient se force ::: $\frac{1}{2} ρ V^{2} S$ se multiply karo $δ$ se $e$ ::: $e = 1/ (1 + δ)$ , penalty $= 1/ e - 1$ Fixed area, span badhao ::: $C_{D, i} \propto S / b^{2}$ , span wins Induced angle ::: $α_{i} = C_{L} / (π A R)$ , aur $C_{D, i} = C_{L} α_{i}$