Aspect ratio — effect on induced drag: Exercises

3.1.23 · D4 · Physics › Compressible Flow & Aerodynamics › Aspect ratio — effect on induced drag

Shuru karne se pehle, har symbol ko phir se samajh lete hain taaki koi bhi reader pehli line se follow kar sake.

Definition Jo symbols hum use karenge — kuch bhi assumed nahi

$b$ = span, wing ki seedhi tip-to-tip distance (ek length, metres mein).
$S$ = planform area, wing ki shadow area seedha upar se dekhi gayi (in $m^{2}$ ).
$c$ = chord, wing ki front-to-back width ek spanwise station par (ek length). Rectangular wing ke liye $S = b c$ .
$A R = b^{2} / S$ = aspect ratio, ek pure number ("span mein kitne chords fit hote hain"). Bada = lamba aur patla).
$C_{L}$ = lift coefficient, ek dimensionless number — "yeh wing lift karne ke liye kitni mehnat kar rahi hai". Slow speed par bada hota hai.
$e$ = span efficiency, ek number $\leq 1$ jo yeh measure karta hai ki lift ki spanwise shape ideal ellipse ke kitni kareeb hai. $e = 1$ perfect hai.
$C_{D, i}$ = induced-drag coefficient, dimensionless drag jo sirf isliye exist karta hai kyunki wing lift banati hai.
$α_{i} = C_{L} / (π A R)$ = induced angle, woh chota angle (radians mein) jis par tip vortices local airflow ko neeche tilt karte hain.
$q = \frac{1}{2} ρ V^{2}$ = dynamic pressure, moving air ka "push"; $ρ$ air density hai, $V$ flight speed hai.

Figure — Aspect ratio — effect on induced drag

Upar ki figure hamaara master reference hai: black bar ek wing hai jo front se dekhi gayi hai, orange curls tip vortices hain, aur red angle $α_{i}$ flow ka downward tilt hai. Har problem is picture ko sahi se padhne ka variation hai.

Level 1 — Recognition

Yahan tumhe sirf sahi formula pick karke plug in karna hai. Arithmetic ke aage koi trap nahi hai.

Problem 1.1

Ek rectangular wing ka span $b = 10 m$ aur constant chord $c = 2 m$ hai. Uska planform area $S$ aur aspect ratio $A R$ nikalo.

Recall Solution 1.1

KYA: Hum $S$ phir $A R$ chahte hain. YEH KYUN: area pehle aata hai kyunki $A R = b^{2} / S$ ko woh chahiye. Rectangle ka area length $\times$ width hota hai: $S = b c = 10 \times 2 = 20 m^{2} .$ Ab aspect ratio: $A R = \frac{b ^{2}}{S} = \frac{1 0 ^{2}}{20} = \frac{100}{20} = 5.$ Check: constant chord ke liye $A R = b / c = 10/2 = 5$ — same answer. Badiya.

Problem 1.2

Words aur symbol mein batao, $C_{D, i}$ ka kya hota hai agar $C_{L}$ aur $e$ fixed rakhein lekin aspect ratio ko triple kar dein.

Recall Solution 1.2

KYA: dependence track karo. $C_{D, i} = C_{L}^{2} / (π e A R)$ mein, $C_{L}, e$ fixed rakhne par, $C_{D, i} \propto 1/ A R$ . $A R$ ko triple karne par denominator 3 se multiply ho jaata hai, isliye: $C_{D, i}^{new} = \frac{1}{3} C_{D, i}^{old} .$ Induced drag ek tihaai ho jaata hai. Yahi "long thin wing wins" wali kahaani hai Aspect ratio — effect on induced drag se.

Problem 1.3

Ek elliptical-loaded wing ka $A R = 8$ hai aur woh $C_{L} = 0.6$ par fly karti hai. $C_{D, i}$ calculate karo.

Recall Solution 1.3

$e = 1$ KYUN: "elliptical loading" ideal case hai, isliye $e = 1$ aur hum short formula use kar sakte hain. $C_{D, i} = \frac{C _{L}^{2}}{π A R} = \frac{0. 6 ^{2}}{π \times 8} = \frac{0.36}{25.13} = 0.01432.$

Level 2 — Application

Ab tumhe formula rearrange karna hoga ya do facts ko chain karna hoga.

Problem 2.1

Ek glider wing ka $A R = 25$ , $e = 0.95$ hai, aur slow thermalling turn mein $C_{L} = 1.0$ produce karna hai. $C_{D, i}$ nikalo.

Recall Solution 2.1

Full formula kyunki $e \neq = 1$ : $C_{D, i} = \frac{C _{L}^{2}}{π e A R} = \frac{1. 0 ^{2}}{π \times 0.95 \times 25} = \frac{1}{74.61} = 0.01340.$ Dhyan do yeh kitna chota hai itne high $C_{L} = 1.0$ ke bawajood: enormous $A R = 25$ kaam kar raha hai. Isliye sailplanes jo dikhte hain woh dikhte hain.

Problem 2.2

Ek wing $C_{L} = 0.8$ aur $e = 1$ par $C_{D, i} = 0.020$ produce karti hai. Uska aspect ratio kya hai?

Recall Solution 2.2

KYA: $A R$ ke liye solve karo. $C_{D, i} = C_{L}^{2} / (π A R)$ se shuru karo aur $A R$ isolate karo: $A R = \frac{C _{L}^{2}}{π C _{D, i}} = \frac{0. 8 ^{2}}{π \times 0.020} = \frac{0.64}{0.06283} = 10.19.$ Sanity check: $A R = 10.19$ wapas plug karo: $0.64/ (π \cdot 10.19) = 0.020$ . ✓

Problem 2.3

Do wings same $C_{L} = 0.7$ aur same $e$ par fly karti hain. Wing A ka $A R = 6$ hai, wing B ka $A R = 15$ hai. B ka induced drag A se kitne factor kam hai?

Recall Solution 2.3

$C_{L}, e$ fixed rakhne par, $C_{D, i} \propto 1/ A R$ , isliye ratio sirf aspect ratios ka inverse ratio hai: $\frac{C _{D, i, B}}{C _{D, i, A}} = \frac{A R _{A}}{A R _{B}} = \frac{6}{15} = 0.40.$ B ka induced drag A ka $0.40 \times$ hai — yaani 60% kam.

Level 3 — Analysis

Yahan tum induced drag ko uske rival, parasite drag ke khilaf weighte karte ho — dekhein Drag polar aur Parasite drag.

Problem 3.1

Total drag coefficient $C_{D} = C_{D, 0} + C_{D, i}$ hai jahan $C_{D, 0} = 0.015$ (roughly constant) parasite part hai. Ek wing ka $A R = 9$ , $e = 0.85$ hai. $C_{L} = 1.1$ par, induced ya parasite drag bada hai, aur total $C_{D}$ kya hai?

Recall Solution 3.1

Induced part: $C_{D, i} = \frac{1. 1 ^{2}}{π \times 0.85 \times 9} = \frac{1.21}{24.03} = 0.05035.$ Compare karo: $C_{D, i} = 0.05035$ vs $C_{D, 0} = 0.015$ . Induced drag bada hai — parasite drag se 3× se zyada — kyunki $C_{L} = 1.1$ high hai (slow flight). Total: $C_{D} = 0.015 + 0.05035 = 0.06535.$

Problem 3.2

Problem 3.1 wali same wing, lekin ab fast cruise mein $C_{L} = 0.3$ par. $C_{D, i}$ aur total dobara compute karo. Ab kaunsa term dominate karta hai?

Recall Solution 3.2

$C_{D, i} = \frac{0. 3 ^{2}}{π \times 0.85 \times 9} = \frac{0.09}{24.03} = 0.003745.$ $C_{D} = 0.015 + 0.003745 = 0.018745.$ Ab parasite drag dominate karta hai ( $0.015$ vs $0.00375$ ). Lesson: induced drag ek low-speed / high- $C_{L}$ problem hai; parasite drag ek high-speed problem hai. Yahi reason hai ki gliders (slow, high $C_{L}$ ) high $A R$ chahte hain jabki fighters (fast, low $C_{L}$ ) nahi chahte — dekhein Glide ratio & L/D max.

Figure dikhata hai ki jaise $C_{L}$ badhta hai do drag pieces kahan cross karte hain: flat gray parasite line aur rising orange $C_{L}^{2}$ induced curve.

Level 4 — Synthesis

Formula ko real dimensional quantities ke saath combine karo: forces, span, area, speed.

Problem 4.1

Ek aircraft ka weight $W = 60, 000 N$ , wing area $S = 30 m^{2}$ , span $b = 20 m$ , $e = 0.9$ hai. Yeh straight aur level fly karta hai (isliye lift $L = W$ ) density $ρ = 1.0 kg/m^{3}$ aur speed $V = 60 m/s$ par. $A R$ , $C_{L}$ , aur actual induced-drag force $D_{i}$ newtons mein nikalo.

Recall Solution 4.1

Step 1 — aspect ratio. $A R = b^{2} / S = 2 0^{2} /30 = 400/30 = 13.33$ . Step 2 — dynamic pressure. $q = \frac{1}{2} ρ V^{2} = 0.5 \times 1.0 \times 6 0^{2} = 0.5 \times 3600 = 1800 Pa$ . Step 3 — lift coefficient. Level flight ka matlab $L = W$ hai, isliye $C_{L} = \frac{L}{q S} = \frac{60000}{1800 \times 30} = \frac{60000}{54000} = 1.111.$ Step 4 — induced-drag coefficient. $C_{D, i} = \frac{C _{L}^{2}}{π e A R} = \frac{1.11 1 ^{2}}{π \times 0.9 \times 13.33} = \frac{1.235}{37.70} = 0.03276.$ Step 5 — coefficient ko force mein wapas convert karo $C_{D, i} = D_{i} / (q S)$ reverse karke: $D_{i} = C_{D, i} q S = 0.03276 \times 1800 \times 30 = 1769 N .$ Toh plane ke total drag mein se, lagbhag $1770 N$ sirf lift banane ki keemat hai.

Problem 4.2

Usi aircraft ke liye, ek designer span ko $b = 26 m$ tak stretch karne ka proposal karta hai ( $S = 30 m^{2}$ aur baaki sab fixed rakhke). Naya induced-drag force kya hai, aur is change se $V = 60 m/s$ par kitni power (in watts) bachti hai? (Power $= D_{i} V$ .)

Recall Solution 4.2

Naya aspect ratio: $A R^{'} = 2 6^{2} /30 = 676/30 = 22.53$ . $C_{L}$ unchanged hai (same $W, q, S$ ) $1.111$ par, isliye $C_{D, i}^{'} = \frac{1.11 1 ^{2}}{π \times 0.9 \times 22.53} = \frac{1.235}{63.71} = 0.01938.$ Naya force: $D_{i}^{'} = 0.01938 \times 1800 \times 30 = 1047 N .$ Power saved: $Δ P = (D_{i} - D_{i}^{'}) V = (1769 - 1047) \times 60 = 722 \times 60 = 43, 320 W \approx 43.3 kW .$ Ek longer, thinner wing yahan 43 kW se zyada engine power bachata hai — real fuel money. (Practice mein tum phir extra wetted area aur structural weight check karte; agla level dekhein.)

Level 5 — Mastery

Full reasoning, optimisation, aur edge cases. Problem statements mein koi hand-holding nahi.

Problem 5.1 — The induced/parasite optimum

Total drag coefficient $C_{D} = C_{D, 0} + \frac{C _{L}^{2}}{π e A R}$ hai. Fixed weight aur speed ke liye (isliye $C_{L}$ fixed hai), induced drag $A R$ ke saath giraata hai lekin parasite drag badhta hai kyunki zyada span matlab zyada wetted area; ise $C_{D, 0} = k A R$ se model karo jahan $k = 0.0016$ hai. $C_{L} = 0.9$ , $e = 0.9$ lo. Woh $A R$ nikalo jo total $C_{D}$ minimise kare, aur minimum $C_{D}$ bhi.

Recall Solution 5.1

KYA: $C_{D} (A R) = k A R + \frac{C _{L}^{2}}{π e} \cdot \frac{1}{A R}$ ko $A R$ ke upar minimise karo. CALCULUS KYUN: function mein ek term $A R$ ke saath badh raha hai aur ek ghaat raha hai; minimum wahan hai jahan unke slopes cancel ho jaayein, yaani jahan derivative zero ho. $C = \frac{C _{L}^{2}}{π e} = \frac{0.81}{π \times 0.9} = \frac{0.81}{2.827} = 0.2865$ likho. Toh $C_{D} = k A R + C / A R$ . Differentiate karo aur zero set karo: $\frac{d C _{D}}{d A R} = k - \frac{C}{A R ^{2}} = 0 \Rightarrow A R^{⋆} = \frac{C}{k} = \frac{0.2865}{0.0016} = 179.1 = 13.38.$ Minimum total drag (note: optimum par dono terms equal hain, har ek $= k A R^{⋆}$ ): $C_{D}^{⋆} = k A R^{⋆} + \frac{C}{A R ^{⋆}} = 2 k A R^{⋆} = 2 \times 0.0016 \times 13.38 = 0.04283.$ Interpretation: sweet spot wahan hai jahan induced drag us extra parasite drag ke barabar ho jaata hai jo use shrink karne mein lagta hai. $A R$ aur push karo aur parasite/weight penalties jeet jaate hain — exactly "higher AR isn't always better" caveat, ab quantitative ban gaya.

Problem 5.2 — Winglet as an effective- $A R$ boost

Ek wing ka geometric $A R = 8$ , $e = 0.80$ hai, $C_{L} = 0.9$ par fly kar rahi hai. Ek winglet effective span efficiency ko $e^{'} = 0.92$ tak badhata hai bina $b$ ya $S$ change kiye. Kyunki $e$ aur $A R$ sirf product $e A R$ ke roop mein enter hote hain, hum ek effective aspect ratio $A R_{eff} = (e^{'} / e) A R$ quote kar sakte hain purane $e$ par. Nikalo: (a) drag reduction percentage mein, aur (b) $A R_{eff}$ .

Recall Solution 5.2

(a) Kyunki $C_{D, i} \propto 1/ e$ baaki sab fixed ke saath: $\frac{C _{D, i}^{'}}{C _{D, i}} = \frac{e}{e ^{'}} = \frac{0.80}{0.92} = 0.8696.$ Yeh $1 - 0.8696 = 0.1304$ ki reduction hai, yaani 13.0% kam induced drag. (b) Product $e A R$ $0.80 \times 8 = 6.40$ se $0.92 \times 8 = 7.36$ ho gaya. Original efficiency $e = 0.80$ par effective aspect ratio ke roop mein express karte hain: $A R_{eff} = \frac{e ^{'} A R}{e} = \frac{0.92 \times 8}{0.80} = 9.2.$ Winglet ek $A R = 8$ wing ko drag-wise aise behave karwata hai jaise $A R = 9.2$ wali ho — physically span badhaaye bina.

Problem 5.3 — The degenerate cases

Formula aur physics use karke explain karo ki $C_{D, i}$ har limit mein kya karta hai: (a) $C_{L} \to 0$ (no lift); (b) $A R \to \infty$ (infinite wing); (c) $e \to 0$ (pathologically bad lift shape). Confirm karo ki har ek parent note ke saath consistent hai.

Recall Solution 5.3

(a) $C_{L} \to 0$ : numerator $C_{L}^{2} \to 0$ , isliye $C_{D, i} \to 0$ . Physics: no lift matlab no pressure difference, no tip leakage, no vortices — kuch bhi flow ko tilt karne wala nahi. Matches "no lift ⇒ no induced drag." (b) $A R \to \infty$ : denominator $\to \infty$ , isliye $C_{D, i} \to 0$ . Physics: infinitely long wing ke tips infinitely door hain — 2-D case jahan induced drag exactly zero hai (dekhein Lifting-line theory (Prandtl)). Yahi reason hai ki induced drag purely ek finite-wing, 3-D effect hai. (c) $e \to 0$ : denominator $\to 0$ , isliye $C_{D, i} \to \infty$ . Physics: $e = 0$ ek impossibly bad, spiky lift distribution hai jo apni saari energy concentrated tip vortices mein dump karti hai. Real wings kabhi uske paas nahi pahunchti ( $e \approx 0.7$ – $1.0$ ), lekin limit sahi warn karti hai ki poor spanwise shape ko bahut penalty milti hai. Best possible case, $e = 1$ , hai Elliptical lift distribution.

Recall Page close karne se pehle ek-line self-test

$C_{D, i}$ formula ::: $C_{D, i} = C_{L}^{2} / (π e A R)$ Coefficient ko force mein convert karo ::: $D_{i} = C_{D, i} q S$ jahan $q = \frac{1}{2} ρ V^{2}$ Optimum $A R$ jab $C_{D, 0} = k A R$ ho ::: $A R^{⋆} = C_{L}^{2} / (π e k)$ , aur wahan dono drags equal hain Teen limits jo zero ya infinite $C_{D, i}$ deti hain ::: $C_{L} \to 0$ (zero), $A R \to \infty$ (zero), $e \to 0$ (infinite)

Connections

Aspect ratio — effect on induced drag — parent formula jo har problem use karti hai
Drag polar — $C_{D} = C_{D, 0} + C_{L}^{2} / (π e A R)$ split jo Levels 3–5 ke peeche hai
Parasite drag — optimisation mein rival term
Wingtip devices (winglets) — Problem 5.2 mein effective- $A R$ idea
Elliptical lift distribution — $e = 1$ ideal
Lifting-line theory (Prandtl) — kyun $A R \to \infty$ induced drag khatam karta hai
Glide ratio & L/D max — jahan drag balance pay off karta hai