Angle of attack, lift coefficient, drag coefficient: Exercises

3.1.20 · D4 · Physics › Compressible Flow & Aerodynamics › Angle of attack, lift coefficient, drag coefficient

Quick symbol reminder (neeche sab kuch inhi par tikaa hai):

Level 1 — Recognition

L1.1 — Dynamic pressure read off karo

$ρ = 1.225 kg/m^{3}$ density ki hawa $V = 50 m/s$ ki speed se flow kar rahi hai. Dynamic pressure $q$ calculate karo.

Recall Solution

KYA: hum seedha definition $q = \frac{1}{2} ρ V^{2}$ mein plug in karte hain. KYUN: $q$ woh akela quantity hai jo density aur speed ko ek saath us "available pressure" mein bundle karta hai jis par saari aerodynamic forces scale hoti hain. $q = \frac{1}{2} (1.225) (50)^{2} = \frac{1}{2} (1.225) (2500) = 1531.25 Pa .$ Toh $q \approx 1531 Pa$ .

L1.2 — Coefficient identify karo

Ek wind-tunnel model $q = 1000 Pa$ ki stream mein $L = 240 N$ lift produce karta hai, reference area $S = 0.3 m^{2}$ hai. $C_{L}$ kya hai?

Recall Solution

KYA: $L = C_{L} q S$ ko rearrange karo taaki dimensionless number nikale. KYUN: $C_{L}$ airstream ( $q$ ) aur size ( $S$ ) ko strip out kar deta hai, sirf pure shape-aur-angle physics rah jaati hai. $C_{L} = \frac{L}{q S} = \frac{240}{1000 \times 0.3} = \frac{240}{300} = 0.80.$ Dimensionless, jaisa hona chahiye (newtons ÷ [Pa · m²] = N ÷ N = 1).

Level 2 — Application

L2.1 — Curve ke linear part par lift

Ek airfoil ka zero-lift angle $α_{L = 0} = - 3^{\circ}$ hai aur lift-curve slope $a_{0} = 0.10 per degree$ hai. $α = 6^{\circ}$ par $C_{L}$ nikalo.

Recall Solution

KYA: $C_{L} = a_{0} (α - α_{L = 0})$ apply karo. KYUN subtraction: slope us angle ko multiply karta hai jo zero-lift line se measure hota hai, chord se nahi. Airfoil $- 3^{\circ}$ par already zero lift bana raha hai, toh effective angle $6 - (- 3) = 9^{\circ}$ hai. $C_{L} = 0.10 (6 - (- 3)) = 0.10 \times 9 = 0.90.$

L2.2 — Coefficient se actual force

Aircraft data: $S = 16 m^{2}$ , $V = 80 m/s$ , $ρ = 1.2 kg/m^{3}$ , $C_{L} = 0.6$ . Lift $L$ newtons mein nikalo.

Recall Solution

Step 1 — dynamic pressure. KYUN: pehle $q$ banao, phir multiply karo. $q = \frac{1}{2} (1.2) (80)^{2} = \frac{1}{2} (1.2) (6400) = 3840 Pa .$ Step 2 — lift. KYUN: $L = C_{L} q S$ matlab coefficient × airstream × area. $L = 0.6 \times 3840 \times 16 = 36, 864 N .$ Sanity check: yeh $36864/9.81 \approx 3758 kg$ support karta hai — ek light aircraft ke liye plausible hai.

L2.3 — Induced drag term

Ek wing $C_{L} = 0.9$ par operate kar raha hai, aspect ratio $AR = 7$ hai, Oswald efficiency $e = 0.85$ hai. Induced-drag coefficient compute karo.

Recall Solution

KYA: $C_{D, i} = \frac{C _{L}^{2}}{π e AR}$ mein plug in karo. KYUN $C_{L}^{2}$ : lift banana flow ko neeche ki taraf bend karta hai (downwash, dekho Induced Drag & Wingtip Vortices), jo lift vector ko peechhe ki taraf tilt karta hai — isse jo drag create hoti hai woh lift ke square ke saath badhti hai. $C_{D, i} = \frac{0. 9 ^{2}}{π ( 0.85 ) ( 7 )} = \frac{0.81}{18.692} = 0.04333.$ Toh $C_{D, i} \approx 0.0433$ .

Level 3 — Analysis

L3.1 — Full drag polar aur $L / D$

Ek wing hai $C_{D, 0} = 0.018$ , $AR = 8$ , $e = 0.9$ , jo $C_{L} = 0.75$ par operate kar raha hai. (a) total $C_{D}$ aur (b) lift-to-drag ratio nikalo.

Recall Solution

Step 1 — induced drag. $C_{D, i} = \frac{0.7 5 ^{2}}{π ( 0.9 ) ( 8 )} = \frac{0.5625}{22.619} = 0.02487.$ Step 2 — total drag. KYUN: profile drag (friction + form) aur induced drag simply add ho jaate hain. $C_{D} = C_{D, 0} + C_{D, i} = 0.018 + 0.02487 = 0.04287.$ Step 3 — glide ratio. KYUN: $L / D = C_{L} / C_{D}$ ( $q S$ factor upar aur neeche cancel ho jaata hai). $\frac{L}{D} = \frac{0.75}{0.04287} = 17.49.$ Toh $C_{D} \approx 0.0429$ aur $L / D \approx 17.5$ .

L3.2 — Same coefficient, speed double

Fixed angle of attack par ek wing ka $C_{L} = 0.5$ hai aur speed $V_{1}$ par $L_{1} = 20, 000 N$ produce karta hai. Agar speed double hokar $2 V_{1}$ ho jaaye (same $α$ , same altitude), toh nayi lift $L_{2}$ kya hogi?

Recall Solution

KYA/KYUN: fixed $α$ par coefficient $C_{L}$ unchanged rehta hai — yeh $α$ , Mach, Reynolds par depend karta hai, directly $V$ par nahi. Sirf $q \propto V^{2}$ change hota hai. Kyunki $L = C_{L} q S$ aur $q \propto V^{2}$ , $\frac{L _{2}}{L _{1}} = \frac{V _{2}^{2}}{V _{1}^{2}} = (2)^{2} = 4.$ $L_{2} = 4 \times 20, 000 = 80, 000 N .$ Lift chaar guni ho jaati hai — isliye nahi ki $C_{L}$ badha, balki isliye ki dynamic pressure badha.

Level 4 — Synthesis

L4.1 — Best glide angle of attack

Ek wing ke liye $C_{D, 0} = 0.020$ , $AR = 10$ , $e = 0.8$ hai, drag polar hai $C_{D} = C_{D, 0} + \frac{C _{L}^{2}}{π e AR}$ . Woh $C_{L}$ nikalo jo $L / D$ ko maximise kare, phir resulting maximum $L / D$ nikalo.

Figure — Angle of attack, lift coefficient, drag coefficient

Recall Solution

KYA chahiye: $\frac{L}{D} = \frac{C _{L}}{C _{D, 0} + k C _{L}^{2}}$ maximise karo jahan $k = \frac{1}{π e AR}$ hai. KYUN derivative: $L / D$ pehle badhta hai (lift drag se tez badhti hai) phir girta hai (induced drag $\propto C_{L}^{2}$ eventually dominate karta hai). Peak wahan hota hai jahan slope zero ho — exactly woh sawaal jo derivative answer karta hai. Figure mein curve dekho: chadta hai, peak karta hai, phir girta hai.

Step 1 — $k$ setup karo. $k = \frac{1}{π ( 0.8 ) ( 10 )} = \frac{1}{25.133} = 0.039789.$ Step 2 — differentiate karo aur zero set karo. Maano $f (C_{L}) = \frac{C _{L}}{C _{D, 0} + k C _{L}^{2}}$ . Quotient rule use karne par, $f^{'}$ ka numerator hai $(C_{D, 0} + k C_{L}^{2}) - C_{L} (2 k C_{L}) = C_{D, 0} - k C_{L}^{2} .$ Isko zero set karne par: $C_{D, 0} = k C_{L}^{2}$ , matlab best glide par induced drag profile drag ke barabar hota hai. $C_{L}^{⋆} = \frac{C _{D, 0}}{k} = \frac{0.020}{0.039789} = 0.50264 = 0.70897.$ Step 3 — wahan drag. Kyunki optimum par $C_{D, i} = C_{D, 0}$ hai, toh $C_{D} = 2 C_{D, 0} = 0.040$ . Step 4 — maximum $L / D$ . $(\frac{L}{D})_{m a x} = \frac{C _{L}^{⋆}}{2 C _{D, 0}} = \frac{0.70897}{0.040} = 17.72.$ Toh best glide $C_{L}^{⋆} \approx 0.709$ par hai aur $L / D_{m a x} \approx 17.7$ hai.

L4.2 — Wind-tunnel model scaling

Ek $1 : 8$ scale model ( $S_{m} = 0.25 m^{2}$ ) ko $q_{m} = 2000 Pa$ par test kiya jaata hai aur $α = 4^{\circ}$ par $L_{m} = 350 N$ milta hai. Full aircraft ( $S = 16 m^{2}$ ) $ρ = 1.0 kg/m^{3}$ , $V = 120 m/s$ par uda, same $α$ (assume karo matched Reynolds & Mach hain toh $C_{L}$ transfer hota hai). Full-scale lift nikalo.

Recall Solution

KYUN coefficient transfer hota hai: $C_{L}$ "shape ki DNA" hai — dimensionless, model aur full scale dono ke liye same $α$ par same (aur matched Reynolds Number, Mach Number ke saath). Isliye coefficients exist karte hain. Step 1 — model coefficient. $C_{L} = \frac{L _{m}}{q _{m} S _{m}} = \frac{350}{2000 \times 0.25} = \frac{350}{500} = 0.70.$ Step 2 — full-scale dynamic pressure. $q = \frac{1}{2} (1.0) (120)^{2} = \frac{1}{2} (14400) = 7200 Pa .$ Step 3 — full-scale lift. $L = C_{L} q S = 0.70 \times 7200 \times 16 = 80, 640 N .$

Level 5 — Mastery

L5.1 — Stall speed aur $C_{L, m a x}$ ceiling

Ek aircraft ka mass $m = 1200 kg$ hai, wing area $S = 18 m^{2}$ hai, aur maximum lift coefficient $C_{L, m a x} = 1.4$ hai (stall se thoda pehle). Sea level par $ρ = 1.225 kg/m^{3}$ , $g = 9.81 m/s^{2}$ lo. Stall speed $V_{s}$ nikalo (woh slowest speed jis par level flight mein lift abhi bhi weight ke barabar ho sake).

Recall Solution

KYA/KYUN: Level flight mein lift = weight: $C_{L} \frac{1}{2} ρ V^{2} S = m g$ . Jab plane slow hota hai, $q$ drop karta hai, toh lift = weight maintain karne ke liye coefficient $C_{L}$ badhna chahiye. Lekin $C_{L}$ $C_{L, m a x}$ se zyada nahi ho sakta (usse aage flow separate ho jaati hai — Boundary Layer Separation & Stall). Toh slowest possible level-flight speed woh hai jahan $C_{L}$ maxed out ho. $C_{L} = C_{L, m a x}$ par setup karo: $C_{L, m a x} \frac{1}{2} ρ V_{s}^{2} S = m g \Rightarrow V_{s} = \frac{2 m g}{ρS C _{L, m a x}} .$ Plug in karo. Numerator: $2 m g = 2 (1200) (9.81) = 23, 544$ . Denominator: $ρS C_{L, m a x} = 1.225 \times 18 \times 1.4 = 30.87$ . $V_{s} = \frac{23544}{30.87} = 762.68 = 27.62 m/s .$ Toh $V_{s} \approx 27.6 m/s$ ( $\approx 99 km/h$ ). Is se neeche, koi bhi angle of attack plane ko upar nahi rakh sakta.

L5.2 — Altitude trade karna: thin air stall speed badhata hai

Wohi aircraft wahan chadh jaata hai jahan $ρ = 0.90 kg/m^{3}$ ho. Stall speed kis factor se change hoti hai, aur nayi $V_{s}$ kya hai?

Recall Solution

KYUN: $V_{s} = 2 m g / (ρS C_{L, m a x})$ se, $ρ$ ke alaawa sab fixed hai, toh $V_{s} \propto 1/ ρ$ . Thin air → zyada stall speed. Factor: $\frac{V _{s, 2}}{V _{s, 1}} = \frac{ρ _{1}}{ρ _{2}} = \frac{1.225}{0.90} = 1.3611 = 1.1667.$ Nayi stall speed: $V_{s, 2} = 1.1667 \times 27.62 = 32.23 m/s .$ Toh stall speed lagbhag $17%$ badhti hai — thin air mein wing ko same lift banane ke liye tez udna padta hai. Isliye high-altitude airfields par zyada lamba take-off run chahiye hota hai.

Connections

Bernoulli's Equation — $q = \frac{1}{2} ρ V^{2}$ ka origin jo har problem mein use hota hai.
Kutta-Joukowski Theorem — L2.1 ke linear lift curve ke peechhe.
Boundary Layer Separation & Stall — L5 mein ceiling $C_{L, m a x}$ .
Induced Drag & Wingtip Vortices — L2.3, L3, L4 mein $C_{L}^{2}$ term.
Reynolds Number & Mach Number — woh conditions jo L4.2 ka coefficient transfer hone deti hain.
Compressible Flow — jahan in coefficients par high-speed corrections aati hain.

Recall Ek-line self-test

Fixed $α$ par speed double karne se lift chaar guni kyun ho jaati hai lekin $C_{L}$ unchanged kyun rehta hai? ::: Kyunki $L = C_{L} q S$ aur $q = \frac{1}{2} ρ V^{2} \propto V^{2}$ hai; coefficient $α$ (aur Mach/Reynolds) par depend karta hai, $V$ par nahi.