Wave drag — transonic and supersonic: Worked examples

3.1.25 · D3 · Physics › Compressible Flow & Aerodynamics › Wave drag — transonic and supersonic

Jo bhi hum use karte hain woh calculation se pehle neeche plain words mein dobara bataya gaya hai, toh yeh page apne aap mein complete hai.

Definition Mach number — master symbol

Mach number $M$ simply flow speed aur local speed of sound ka ratio hai: $M = \frac{v}{a},$ jahan $v$ woh speed hai jis par hawa (ya body hawa ke relative) move kar rahi hai aur $a$ woh speed hai jis par pressure signals us hawa mein travel karti hain. $M$ ek pure number hai (ek speed doosri speed se divided). Is page par do flavours aate hain:

$M_{\infty}$ (free-stream Mach): body se door undisturbed flight speed use karta hai — "plane kitni tez udd raha hai."
$M_{l oc a l}$ (local Mach): surface ke kisi particular point par sped-up flow use karta hai (hawa ek curved wing par accelerate hoti hai). Kyunki hawa locally tez hoti hai, $M_{l oc a l} > M_{\infty}$ . Jab bhi is page par $M^{2} - 1$ dikhega, $M$ ka matlab $M_{\infty}$ hai (ek whole-airfoil property) jab tak hum explicitly $M_{l oc a l}$ na likhen.

Formula

$θ$ – $β$ – $M$ relation (yahan stated, Ex 4 & 5 mein use hoga) Mach $M_{1}$ par ek supersonic flow ke liye jo ek surface se milti hai jo use angle $θ$ se moddti hai, ek attached oblique shock tilt $β_{shock}$ par banta hai. Shock ke across mass, momentum aur energy ka conservation deta hai: $tan θ = 2 cot β_{shock} \frac{M _{1}^{2} s i n ^{2} β _{shock} - 1}{M _{1}^{2} ( γ + c o s 2 β _{shock} ) + 2} .$ Kaise padhein: turn $θ$ aur incoming $M_{1}$ pata hai; shock tilt $β_{shock}$ solve karte hain. Yaad rakhne ki do cheezein:

Jaise $θ \to 0$ (vanishingly weak turn) weak-shock root $β_{shock} \to μ = arcsin (1/ M_{1})$ — Mach angle. Yahi woh only case hai jahan ek real shock Mach cone ke equal hoti hai.
Har $M_{1}$ ke liye ek maximum $θ_{ma x}$ hota hai. Isse zyada turning maango toh koi attached shock exist nahi karti — yeh detach ho jaati hai (Ex 5). Hum is page par kabhi isse re-derive nahi karenge; hum sirf use karenge (numerical roots standard tables/solver se aate hain air ke liye, $γ = 1.4$ ).

Definition Do aur symbols jinhein hum use karenge

Mach angle $μ$ : Mach cone ka half-angle, woh cone jo weakest possible pressure signals sweep karta hai jab ek body sound se tez chalti hai. Parent ki pure geometry se (pulse radius $a t$ vs distance travelled $u t$ ), $sin μ = 1/ M$ , toh $μ = arcsin (1/ M)$ . Yeh ek zero-strength wave ka tilt hai — hamesha flow mein shallowest line.
Ratio of specific heats $γ$ : gas ki ek property — constant pressure par uski heat capacity aur constant volume par uski heat capacity ka ratio. Ordinary air ke liye (ek diatomic gas: mostly $N_{2}$ aur $O_{2}$ ) $γ = 1.4$ hai. Har shock relation mein $γ$ hota hai kyunki gas ko compress karna use garam karta hai, aur $γ$ measure karta hai kitna strongly. Hum poore mein $γ = 1.4$ use karte hain kyunki working fluid air hai.
Mean-square slope $\overline{(d y / d x)^{2}}$ : overline ka matlab hai "chord ke along average." Tum surface slope $d y / d x$ leading edge se trailing edge tak har point par lete ho, use square karte ho, phir un sab squares ko chord length par average karte ho. Yeh ek single number hai jo summarise karta hai ki surface kitni steep hai, average mein — aur pehle square karna matlab hai ki up-slopes aur down-slopes dono drag add karte hain, kabhi cancel nahi karte.

Scenario matrix

Har wave-drag problem in cells mein se kisi ek mein aata hai. Neeche worked examples un cell(s) ke saath label hain jo woh cover karte hain, toh saath mein poora grid fill hota hai.

Cell	Case class	Physics kaun decide karta hai	Example(s)
A	$M_{\infty} < 1$ , har jagah subsonic	koi shock nahi ⇒ koi wave drag nahi	Ex 1
B	$M_{\infty} < 1$ lekin local pocket $M = 1$ hit karta hai	onset — $M_{cr}$ define karta hai	Ex 2
C	$M_{\infty} = 1$ exactly (degenerate)	$β_{P G} = 0$ ⇒ formulas blow up	Ex 3
D	$M_{\infty} > 1$ , slender body, attached shock	$θ < θ_{ma x}$ , real $β_{shock} > μ$	Ex 4
E	$M_{\infty} > 1$ , blunt/over-turned, detached bow shock	$θ > θ_{ma x}$	Ex 5
F	Supersonic thin airfoil, lifting	Ackeret $c_{l}, c_{d}$ with $α \neq = 0$	Ex 6
G	Supersonic thin airfoil, zero lift	sirf thickness/camber term	Ex 7
H	Sign / limiting behaviour	$c_{d} \geq 0$ hamesha; $M \to \infty$ trend	Ex 7, Ex 8
I	Real-world word problem	sweep, $M_{n} = M_{\infty} cos Λ$	Ex 8
J	Exam twist — do $β$ 's	symbol clash pakdo	Ex 9

Example 1 — Cell A: subsonic, koi wave drag nahi

Worked example Statement

Ek wing $M_{\infty} = 0.5$ par fly kar raha hai. Uska fastest local point $M_{l oc a l} = 0.82$ tak pahunchta hai. Kitna wave drag banata hai?

Forecast: Kya flow ka koi hissa supersonic hai? Agar nahi, toh yeh wave drag ko kya force karta hai?

Kahin bhi sabse bada local Mach number dhundho. Diya gaya hai: $M_{l oc a l} = 0.82$ . Yeh step kyun? Wave drag tab hi exist karta hai jab flow kahin $M \geq 1$ reach kare. Toh poora sawaal yeh reduce hota hai ki "kya koi point sonic hit karta hai?"
1 se compare karo. $0.82 < 1$ , toh flow har jagah subsonic hai. Koi supersonic pocket nahi ⇒ koi shock form nahi ho sakta. Yeh step kyun? Koi shock nahi matlab koi entropy jump nahi, koi stagnation-pressure loss nahi, koi wave drag nahi — parent mein definition ke hisaab se.
Conclude: $c_{d, wave} = 0$ .

Verify: Kahin bhi sabse bada Mach number $M_{l oc a l} = 0.82 < 1$ hai. Toh fastest point bhi kabhi sonic nahi pahunchta, supersonic factor $M_{l oc a l}^{2} - 1$ imaginary hoga, aur supersonic machinery simply apply nahi hoti. Answer $0$ consistent hai.

Example 2 — Cell B: onset, $M_{cr}$ conceptually dhundhna

Worked example Statement

Kis rule se hum critical Mach number $M_{cr}$ locate karte hain, aur ek thicker airfoil ka $M_{cr}$ lower kyun hota hai?

Forecast: $M_{cr}$ par fastest point par local Mach number kya hoga — exact value guess karo.

Defining condition state karo: airfoil ke minimum-pressure (maximum-velocity) point par $M_{l oc a l} = 1$ set karo. Yeh step kyun? $M_{cr}$ define hai woh free-stream $M_{\infty}$ ke roop mein jis par sonic flow pehli baar kahin appear hoti hai. Sonic hone wali pehli jagah fastest point hoga.
Local ko free-stream se link karo: local speed-up factor $k = M_{l oc a l} / M_{\infty} > 1$ geometry par depend karta hai. Thicker section ⇒ hawa ko bulge ke upar se jaane ke liye aur accelerate karna padta hai ⇒ bada $k$ . Yeh step kyun? Bada $k$ matlab local flow chhote $M_{\infty}$ par $M = 1$ reach karta hai.
Conclude: bada $k$ ⇒ $M_{cr} = 1/ k$ (schematically) chhota hai. Thick wings pehle trouble mein padti hain.

Verify: Parent ke design rule ke saath consistent — high-speed wings thin aur swept hoti hain exactly $k$ chhota aur $M_{cr}$ zyada rakhne ke liye. Logic direction (thicker → lower $M_{cr}$ ) match karta hai.

Example 3 — Cell C: degenerate case $M_{\infty} = 1$

Worked example Statement

Ackeret lift formula $c_{l} = \frac{4 α}{M _{\infty}^{2} - 1}$ exactly $M_{\infty} = 1$ par use karne ki koshish karo. Kya hota hai, aur physically kya matlab hai?

Forecast: Denominator dekho. $M = 1$ par kya hota hai?

$M_{\infty} = 1$ substitute karo: $1^{2} - 1 = 0 = 0$ . Yeh step kyun? Poora formula Prandtl–Glauert factor $β_{P G} = M^{2} - 1$ par bana hai; uski edge test karne se pata chalta hai theory kahan marti hai.
Isse divide karo: $c_{l} = 4 α /0 \to \infty$ . Yeh step kyun? Prediction infinite hai — ek clear signal ki linearized theory $M = 1$ par invalid hai.
Physically interpret karo: $M = 1$ par Mach angle $μ = arcsin (1/1) = 90°$ ; disturbances flow ke perpendicular pile up hoti hain, small-perturbation assumption collapse hoti hai, aur real drag finite hai lekin formula nahi. Yeh step kyun? Math singularity real transonic drag spike ko mirror karti hai — woh regime jahan full nonlinear (ya experimental) methods use karne padte hain, Ackeret nahi.

Verify: $μ = 90°$ at $M = 1$ aur $μ \to 0$ as $M \to \infty$ ( $sin μ = 1/ M$ se), toh singularity exactly wahin baithti hai jahan physics sabse mushkil hai — internally consistent.

Example 4 — Cell D: attached oblique shock, real $β > μ$

Neeche wala figure ek $M_{1} = 2$ flow dikhata hai jo ek $10°$ half-angle wedge (grey) se takra raha hai, head-on drawn. Wedge tip se teen lines nikalti hain: black dashed Mach line $μ = 30°$ par (weakest signal), aur red real shock $β_{shock} = 39.3°$ par (steeper). Incoming flow arrow left se enter karta hai. Figure ka point: red shock hamesha dashed Mach line se zyada lean karta hai.

Figure — Wave drag — transonic and supersonic

Worked example Statement

Ek $10°$ half-angle wedge $M_{1} = 2.0$ par fly karta hai. Mach angle $μ$ (weak signals ka envelope) aur real attached shock angle $β_{shock}$ compare karo.

Forecast: Dono mein se kaun bada hai — weak-signal cone, ya woh actual shock jo flow ko $10°$ se turn karta hai?

Mach angle: $sin μ = 1/ M_{1} = 1/2 \Rightarrow μ = arcsin (0.5) = 30°$ . Yeh step kyun? $μ$ pure geometry hai "pulse radius $a t$ vs distance $u t$ " ki — weakest possible disturbance. Yeh hamara lower bound hai.
Real shock angle: $\theta$–$\beta$–$M$ relation (upar box mein stated) solve karo $θ = 10°$ , $M_{1} = 2$ , $γ = 1.4$ (air) ke saath. Weak-shock root $β_{shock} \approx 39.3°$ hai. Yeh step kyun? Ek finite $10°$ turn ko finite compression chahiye, aur finite compression matlab shock zero-strength Mach cone se zyada lean karta hai.
Compare karo: $39.3° > 30°$ . Real shock steeper hai — figure mein red shock line upar black dashed Mach line se baithti hai. Yeh step kyun? Yeh parent ki key caution hai jo concrete banayi gayi hai — kabhi real shock angle ko $μ$ ke barabar mat set karo.

Verify: Jaise $θ \to 0$ , $θ$ – $β$ – $M$ weak root $β = μ = 30°$ par relax honi chahiye. $β = 30°$ , $M_{1} = 2$ numerator mein plug karo $M_{1}^{2} sin^{2} β - 1 = 4 (0.25) - 1 = 0$ , toh $θ = 0$ . ✓ Consistent — Mach angle exactly zero-turn shock hai.

Example 5 — Cell E: over-turning ⇒ detached bow shock

Neeche wala figure same $M_{1} = 2$ flow dikhata hai jo bahut blunter wedge (grey, half-angle $40°$ ) se takra raha hai. Kyunki wedge attached shock se zyada turning demand karta hai, shock (red curve) tip touch nahi kar sakti — yeh bow out hoti hai aur body se aage khadhi hoti hai, ek visible stand-off gap red curve aur wedge tip ke beech marked hai. Incoming flow arrow left se enter karta hai.

Worked example Statement

Ek wedge with bahut bada half-angle $θ = 40°$ ek $M_{1} = 2.0$ flow mein rakhi gayi hai. Kya shock attached reh sakti hai?

Forecast: Kya $40°$ Mach 2 ke liye ek gentle turn hai ya violent wala? Guess karo shock tip se chipki rahegi ya aage khadhi ho jaayegi.

$M_{1} = 2$ ke liye maximum turning angle dhundho. $θ$ – $β$ – $M$ relation se ( $γ = 1.4$ air ke liye), $θ_{ma x} \approx 22.97°$ at $M_{1} = 2$ . Yeh step kyun? Har Mach number ke liye ek hard ceiling hoti hai ki ek attached shock flow ko kitna turn kar sakti hai. Isse aage, koi attached solution exist nahi karta.
Compare karo: requested turn $40° > θ_{ma x} \approx 23°$ . Yeh step kyun? Body se zyada turning maanga ja raha hai jo koi bhi attached oblique shock provide kar sake.
Conclude: shock wedge se aage khadhi ek curved bow shock mein detach ho jaati hai (figure mein red curve). Centreline ke paas yeh locally ek normal shock hai — sabse strong, sabse zyada entropy generate karne wali. Yeh step kyun? Stronger shock ⇒ bada $Δ s$ ⇒ bada stagnation-pressure loss ⇒ maximum wave drag. Exactly isliye blunt/over-angled bodies draggy hoti hain.

Verify: $θ_{ma x} (M = 2) \approx 23°$ aur hamara turn $40°$ hai; kyunki $40 > 23$ "no attached solution" verdict forced hai. $M_{1} = 2$ par normal shock stagnation-pressure ratio $p_{0, 2} / p_{0, 1} \approx 0.721$ deta hai — ek bada loss, high drag confirm karta hai. ✓

Example 6 — Cell F: lifting supersonic flat plate

Worked example Statement

Ek flat plate $M_{\infty} = 2.0$ par, angle of attack $α = 3° = 0.0524$ rad. $c_{l}$ , $c_{d, wave}$ , aur $L / D_{wave}$ dhundho.

Forecast: Kya $L / D$ roughly 2, 20, ya 200 ke aas-paas hoga? (Hint: yeh roughly $1/ α$ hai.)

Prandtl–Glauert factor compute karo: $β_{P G} = M_{\infty}^{2} - 1 = 4 - 1 = 3 = 1.732$ . Yeh step kyun? Har Ackeret result $β_{P G}$ se divided hota hai; yeh supersonic (shock-angle wale sense mein nahi) ke liye compressibility strength hai.
Lift coefficient: $c_{l} = \frac{4 α}{β _{P G}} = \frac{4 ( 0.0524 )}{1.732} = 0.1210$ . Yeh step kyun? Ackeret ki linear theory kehti hai lift tilt $α$ ke saath linearly badhti hai — oblique-wave pressure ka small- $θ$ limit.
Wave drag (flat plate, koi thickness/camber nahi): $c_{d, wave} = \frac{4 α ^{2}}{β _{P G}} = \frac{4 ( 0.0524 ) ^{2}}{1.732} = 0.006340$ . Yeh step kyun? Flat plate ke liye sirf lift-induced wave-drag term $α^{2}$ bachta hai (koi thickness, koi camber nahi).
Efficiency: $L / D_{wave} = c_{l} / c_{d, wave} = \frac{4 α}{4 α ^{2}} = \frac{1}{α} = \frac{1}{0.0524} = 19.1$ . Yeh step kyun? $β_{P G}$ aur $4$ cancel ho jaate hain, $L / D_{wave} = 1/ α$ is linearized theory mein bachta hai. Flat plate ka exact geometric result $L / D = cot α$ hai; kyunki $cot α = cos α / sin α \approx 1/ α$ sirf tab jab $α$ small ho (toh $sin α \approx α$ , $cos α \approx 1$ ), $1/ α$ form small $α$ ke liye valid approximation hai. $α = 3°$ par yeh $0.2%$ se better agree karte hain.

Verify: $c_{l} / c_{d} = 0.1210/0.006340 = 19.1$ aur $1/0.0524 = 19.1$ . Exact $cot (3°) = 19.08$ — $1/ α = 19.1$ se $0.1%$ ke andar, small-angle approximation confirm karta hai. ✓ Units: saare coefficients dimensionless, ratio dimensionless. Small $α$ efficient hai — lekin absolute lift $0.1210$ phir small bhi hoti hai, supersonic eternal trade-off.

Example 7 — Cells G & H: zero-lift thickness drag, aur "kabhi negative nahi" fact

Worked example Statement

Ek symmetric diamond (double-wedge) airfoil zero lift par $M_{\infty} = 2$ par fly karta hai ( $α = 0$ , koi camber nahi). Uski thickness slope ka mean-square value $\overline{(d y_{t} / d x)^{2}} = 0.01$ hai. $c_{d, wave}$ dhundho aur confirm karo ki yeh negative nahi ho sakta.

Forecast: Koi lift nahi aur koi camber nahi — kya wave drag zero hai? Ya sirf volume hona bhi drag cost karta hai?

Lift aur camber terms drop karo. $α = 0$ aur $\overline{(d y_{c} / d x)^{2}} = 0$ , toh sirf thickness term bachti hai: $c_{d, wave} = \frac{4}{β _{P G}} \overline{(d y_{t} / d x)^{2}} .$ Yeh step kyun? Parent ka Ackeret drag mein teen additive squares hain; lift aur camber zero karne se volume (thickness) contribution isolate hota hai.
Numbers daalo: $β_{P G} = 3 = 1.732$ , toh $c_{d, wave} = \frac{4 ( 0.01 )}{1.732} = 0.0231$ . Yeh step kyun? Dikhata hai ki wave drag zero lift par bhi nonzero hai — pure volume shocks banata hai.
Sign argument (Cell H). Har term ek square hai ( $α^{2}, \overline{(\cdot)^{2}}$ ) aur $β_{P G} > 0$ , toh $c_{d, wave} \geq 0$ hamesha — yeh kabhi negative nahi ho sakta, aur minimize hota hai (→0) sirf infinitely thin, uncambered plate par zero lift ke saath. Yeh step kyun? Wave drag ek loss hai (har shock ke across entropy badhti hai); negative drag matlab shock return karti hai energy body ko, jo second law forbid karta hai.

Verify: $4 (0.01) /1.732 = 0.0231$ . ✓ Squares non-negativity guarantee karte hain: kisi bhi slope ka sign flip karne par bhi square positive rehta hai. Entropy-only-rises ke saath consistent.

Example 8 — Cells H & I: $M \to \infty$ trend, aur drag divergence se bachne ke liye sweep

Worked example Statement

Part (a) — limiting behaviour (Cell H): Flat-plate wave drag $c_{d, wave} = \frac{4 α ^{2}}{M _{\infty}^{2} - 1}$ lo fixed $α$ par. $M_{\infty} \to \infty$ par kya hota hai, aur $M_{\infty} \to 1^{+}$ par? Part (b) — word problem (Cell I): Ek airliner $M_{\infty} = 0.85$ par cruise karta hai. Uski unswept wing section ka $M_{cr} = 0.72$ hoga (drag already diverging). Designer wing ko $Λ = 35°$ sweep karta hai. Wing section actually konsa normal Mach number "feel" karta hai, aur kya yeh drag rise fix karta hai?

Forecast: (a) Kya zyada tez supersonically jaane se wave drag badhta hai ya ghatta hai? (b) Kya $M_{n} = M_{\infty} cos Λ$ $0.72$ se neeche jayega?

(a) High-Mach limit. Jaise $M_{\infty} \to \infty$ , $M_{\infty}^{2} - 1 \to \infty$ , toh $c_{d, wave} = \frac{4 α ^{2}}{M _{\infty}^{2} - 1} \to 0$ . Yeh step kyun? Akela $M$ -dependence denominator mein hai; growing denominator coefficient ko drive down karta hai. Physically, oblique shocks tighter lean karti hain aur per unit length weaker hoti hain, toh wave-drag coefficient increasing supersonic speed ke saath girta hai.
(a) Low-Mach (transonic) limit. Jaise $M_{\infty} \to 1^{+}$ , $M_{\infty}^{2} - 1 \to 0^{+}$ , toh $c_{d, wave} \to + \infty$ . Yeh step kyun? Yeh same singularity hai jaise Ex 3 mein — linearized theory $M = 1$ ke paas over-predict karti hai, real transonic drag spike mirror karte hue. Dono limits ke beech, $c_{d, wave}$ $M_{\infty} > 1$ ke liye monotonically girta hai. Sample numbers $α = 0.0524$ par: $M = 1.4 \Rightarrow \frac{4 α ^{2}}{0.96} = 0.01121$ ; $M = 2 \Rightarrow 0.006340$ ; $M = 5 \Rightarrow \frac{4 α ^{2}}{24} = 0.002242$ — steadily shrinking. ✓ trend.
(b) Normal Mach number compute karo: $M_{n} = M_{\infty} cos Λ = 0.85 \times cos 35° = 0.85 \times 0.8192 = 0.696$ . Yeh step kyun? Sirf woh velocity component jo wing ke leading edge ke perpendicular hai section par pressure field drive karta hai, toh woh component local sonic onset set karta hai — full flight speed nahi.
(b) Section $M_{cr} = 0.72$ se compare karo: $0.696 < 0.72$ . Yeh step kyun? Agar felt Mach number critical se neeche rahe, section par koi supersonic pocket nahi banti ⇒ wave drag suppress hota hai even though plane $0.85$ par fly kar raha hai.
(b) Limiting check. Zyada sweep ⇒ chhota $cos Λ$ ⇒ chhota $M_{n}$ ⇒ divergence se pehle higher effective cruise Mach. Jaise $Λ \to 90°$ , $M_{n} \to 0$ (lekin wing lift ke liye useless ho jaati hai — practical trade-off).

Verify: (a) $M = 5$ : $4 (0.0524)^{2} / 24 = 0.002242 < 0.006340$ ( $M = 2$ value) $< 0.01121$ ( $M = 1.4$ value) — monotone decrease confirmed. (b) $0.85 cos 35° = 0.696 < 0.72$ ✓ — sweep margin wapas dilata hai toh section cruise par supersonic nahi hoti. Saare Mach numbers dimensionless.

Example 9 — Cell J: exam twist — ek problem mein do $β$ 's

Worked example Statement

$M_{\infty} = 2.0$ par: ek student likhta hai "shock angle $β = M^{2} - 1 = 1.732$ hai." Kya galat hai, aur woh do correct quantities konsi hain jinke beech yeh ek symbol confuse ho raha hai?

Forecast: Kya koi physical angle bare number $1.732$ ke barabar ho sakta hai? (Sochte hain angle ki units ke baare mein vs pure number.)

Category error spot karo. $M^{2} - 1 = 4 - 1 = 1.732$ ek dimensionless number hai, Prandtl–Glauert factor $β_{P G}$ — yeh angle nahi hai. Angle ke roop mein padha jaaye toh yeh ya toh $1.732$ rad $= 99.2°$ hoga (Mach-2 shock ke liye absurd) ya fir koi units nahi hogi. Yeh step kyun? Student ne opening definition box ke $β$ ke do meanings collide kar diye — exactly woh trap jo is cell pakadne ke liye exist karta hai.
Real shock angle do. Ek concrete body ke liye — maan lo $10°$ wedge — physical shock tilt $θ$ – $β$ – $M$ relation se $β_{shock} \approx 39.3°$ hai (Example 4). Yeh step kyun? Real shock angle shock relations se aata hai aur turning angle $θ$ par depend karta hai; yeh kabhi $M^{2} - 1$ nahi hota.
Contrast ke liye Mach angle do. $μ = arcsin (1/ M) = arcsin (0.5) = 30°$ . Yeh step kyun? Yeh nail down karta hai ki $M = 2$ par teen distinct numbers hain: $β_{P G} = 1.732$ (Ackeret denominator mein ek pure number), $μ = 30°$ (weak-signal cone angle), aur $β_{shock} = 39.3°$ ( $10°$ wedge ke liye real shock tilt). Inhe alag mental boxes mein rakho.

Verify: $4 - 1 = 1.732$ (ek number); $arcsin (0.5) = 30°$ ; $10°$ wedge shock $\approx 39.3°$ . Teen alag values teen alag meanings ke saath — aur $1.732 rad = 99.2° \neq = 39.3°$ , prove karta hai $β_{P G}$ shock angle nahi hai. ✓

Recall Self-test

Mach number ek line mein define karo. ::: $M = v / a$ , flow speed aur local speed of sound ka ratio (ek pure number). Kahin bhi koi shock nahi matlab wave drag kya hai? ::: Exactly zero (Cell A). Exactly $M = 1$ par $c_{l} = 4 α / M^{2} - 1$ kyun fail karta hai? ::: Denominator $\to 0$ , prediction $\to \infty$ ; linearized theory $M = 1$ par invalid hai (Cell C). $M = 2$ par $10°$ wedge ke liye $μ$ , $β_{shock}$ order karo, aur kya $β_{P G}$ belong karta hai? ::: $μ = 30° < β_{shock} = 39.3°$ ; $β_{P G} = 1.732$ ek pure number hai, angle nahi (Cells D, J). Oblique shock bow shock mein kab detach hota hai? ::: Jab required turn $θ$ , $θ_{ma x}$ se zyada ho jaaye (≈ $23°$ at $M = 2$ ) — Cell E. $c_{d, wave}$ kabhi negative kyun nahi ho sakta? ::: Har term ek square hai aur $M^{2} - 1 > 0$ ; negative wave drag entropy-rise violate karega (Cell H). $M_{\infty} \to \infty$ par fixed $α$ ke saath $c_{d, wave}$ kya karta hai? ::: Yeh zero ki taraf girta hai $1/ M^{2} - 1$ ki tarah (Cell H). Swept wing section konsa Mach number "feel" karta hai? ::: Normal component $M_{n} = M_{\infty} cos Λ$ (Cell I).