Wave drag — transonic and supersonic: Visual walkthrough

3.1.25 · D2 · Physics › Compressible Flow & Aerodynamics › Wave drag — transonic and supersonic

Step 1 — Ek chota sa sound pulse sirf ek spreading ring hai

KYA. Imagine karo ki still paani mein ek patthar giraya. Ek ring fixed speed par bahar ki taraf spread hoti hai. Air bhi aisa hi karta hai pressure ke saath: koi bhi choti si thapki ek pressure ring bahar bhejti hai. Us speed ko speed of sound kehte hain, jo $a$ likhte hain.

HUM YAHAN SE KYUN SHURU KARTE HAIN. Air ko pata chalta hai ki kuch aa raha hai sirf inhi pressure rings se. Agar rings body se pehle aage ki air tak pahunch jaati hain, toh air ko dhire se side mein hatne ka time milta hai. Agar nahi pahunchti, toh air surprise ho jaati hai — aur surprise wahi hai jahan shocks (aur drag) janam lete hain. Toh poori kahani ek race hai body aur uske apne rings ke beech.

PICTURE. Ek single point ek pulse create karta hai. Time $t$ ke baad wo pulse ek circle mein grow ho chuka hai jiska radius $a t$ hai — speed times time, simple "distance = speed × time" rule.

Step 2 — Bahut saari pulses, aur woh number jo sab kuch decide karta hai

KYA. Body air ko ek baar nahi pokti — wo continuously pokti hai jab move karti hai. Toh har instant par wo ek nayi expanding ring drop karti hai, jabki khud aage slide karti rehti hai. Hum compare karte hain, same time $t$ mein:

ring kitni far spread hui: $a t$ ,
body kitni door gayi: $u t$ .

KYUN. Unka ratio hi ek ek cheez hai jo matter karta hai, kyunki picture identical dikhti hai chahe $t$ ek second ho ya ek saal — sirf ratio $u / a$ geometry ko change karta hai. Hum us ratio ko uska apna naam dete hain, Mach number:

$M = body speed u / ring speed a .$

$M < 1$ : body apni rings se slow → rings aage nikal jaati hain → air warned ho jaati hai.
$M = 1$ : body apni rings ke front ke saath pace rakhti hai.
$M > 1$ : body apni rings ko outrun karti hai → aage ki air kabhi warned nahi hoti.

PICTURE. Teen panels: rings aage escape karti huin (subsonic), rings sirf body ko touch karti huin (sonic), rings peeche reh kar pile up karti huin (supersonic).

Recall Kyun ratio, raw speeds nahi?

Kyun $M = u / a$ , aur sirf " $u$ " nahi, flow decide karta hai? ::: Kyunki body-vs-rings ki geometry sirf is baat par depend karti hai ki do distances $u t$ aur $a t$ kaise compare hoti hain — unka ratio. Same $M$ ⇒ same picture, kisi bhi scale par.

Step 3 — Jab body apni rings ko outrun karti hai: the Mach cone

KYA. Supersonic case lo ( $M > 1$ ). Body ne apne path mein jo bhi rings emit ki hain unhe draw karo. Kyunki body har ring ke leading edge se aage hai, saari rings ek straight line ke neeche neat tuck ho jaati hain jo unhe sab ko touch karti hai — poori family ki ek tangent. 3-D mein woh tangent line ek cone mein sweep karti hai: Mach cone.

CONE KYU MATTER KARTA HAI. Cone ke bahar ki air ne kuch bhi feel nahi kiya — total silence, koi warning nahi. Andar, pulses aa chuke hain. Cone exactly woh boundary hai "warned" aur "un-warned" air ke beech. Iska opening batata hai ki supersonic flow kitni door tak pahunchti hai; dekho Sonic boom and the Mach cone.

PICTURE — angle kahan se aata hai. Ek purani ring se ek right triangle banao:

ring ka radius $a t$ woh side hai jo cone ke half-angle $μ$ ke opposite hai,
body ka travel $u t$ hypotenuse hai (straight-line distance wahan se jahan woh ring paida hui thi, wahan tak jahan body ab hai).

Kisi angle ka sine hota hai opposite over hypotenuse — aur yahi exactly woh ratio hai jo humein chahiye:

$sin μ = \frac{opposite}{hypotenuse} = \frac{a t}{u t} = \frac{a}{u} = \frac{1}{M} .$

Saare cases (ek bhi skip mat karo):

$M$	$1/ M$	$μ = arcsin (1/ M)$	Matlab
$\to 1^{+}$	$\to 1$	$\to 90°$	Cone flat khulta hai — barely supersonic
$2$	$0.5$	$30°$	Ek clean example
$\to \infty$	$\to 0$	$\to 0°$	Cone body ke tight wrap ho jaata hai
$< 1$	$> 1$	undefined	$arcsin$ of a number $> 1$ ka koi jawab nahi — koi cone exist nahi karta, jo "rings escape ahead" se match karta hai

Step 4 — Kyun silence-then-arrival ek shock force karta hai

KYA. Supersonic flow mein ek slender body ki nose ke paas zoom karo. Bilkul aage ki air "silent" zone ke andar hai — usne kuch nahi suna, toh wo abhi bhi full speed par seedhi flow ho rahi hai. Ek hair's-width baad use suddenly body ke around bend karna padega. Nature ke paas koi smooth ramp available nahi hai (warning kabhi aayi hi nahi), toh change ek razor-thin front ke across hota hai: ek shock wave.

WOH ABRUPT KYUN HAI. Subsonic flow mein, pressure rings aage bhaagte hue air ko apni nayi direction mein ease karne dete hain ek lambi distance par. Woh early warning hato (supersonic) aur poora turn lagbhag-zero thickness mein cram ho jaata hai. Badi change chhoti distance mein = ek shock. Dekho Normal and oblique shock waves.

PICTURE. Left: subsonic, streamlines dhire aur pehle curve karti hain. Right: supersonic, streamlines bilkul seedhi chalti hain phir ek shock line par sharply kink ho jaati hain.

Step 5 — Ek real shock Mach cone se ZYADA lean karta hai

KYA. Half-angle $θ$ ka ek wedge (flow ko kitna turn karna hai) ek $M_{1} > 1$ stream mein rakho. Attached shock angle $β$ par baithta hai, aur $β$ shock ke across mass/momentum/energy books se fix hota hai — == $θ$ – $β$ – $M$ relation==:

$tan θ = 2 shock lean cot β \frac{M _{1}^{2} sin ^{2} β - 1 normal Mach > 1 \Rightarrow real shock}{M _{1}^{2} ( γ + c o s 2 β ) + 2} .$

$θ$ = wall flow ko kitna hard turn karti hai (woh cause).
$β$ = shock ka tilt (woh effect jiske liye hum solve karte hain). Note: is step mein $β$ ka matlab hai oblique-shock angle; yahi letter Step 7 mein alag quantity ke liye reuse hota hai — hum wahan flag karenge.
$M_{1}$ = upstream Mach number.
$γ$ = ratio of specific heats ( $\approx 1.4$ air ke liye) — gas kitna "springy" hai.
$M_{1} sin β$ = woh Mach number jo shock ke across measure hota hai; real shock ke liye ise $1$ se exceed karna hoga, jo $β > μ$ force karta hai.

KYUN $β > μ$ HAMESHA. $θ \to 0$ set karo (ek vanishingly gentle turn). Tab shock strength kuch nahi ho jaati aur $β \to μ = arcsin (1/ M_{1})$ — Mach cone zero-strength limit hai. Koi bhi real turn ( $θ > 0$ ) real compression chahta hai, toh $β$ $μ$ se upar chadh jaata hai. Dekho Theta-beta-M relation and detached bow shocks.

PICTURE. Mach cone (dashed, angle $μ$ ) ek wedge nose se steeper solid shock (angle $β$ ) ke neeche draw kiya gaya hai.

Degenerate case — shock detach ho jaata hai. Har $M_{1}$ ke liye ek sabse bada turn $θ_{ma x}$ hota hai. Usse zyada turn maango (ek blunt nose), aur koi attached shock exist nahi karta: wo body se pop off ho jaata hai aur ek curved bow shock ki tarah aage khada ho jaata hai, centreline par locally normal (uska steepest, strongest form). Blunt = strongest shock = most wave drag.

Step 6 — Kyun ek shock IS drag hai: the entropy ledger

KYA. Ek shock total energy conserve karta hai (yeh fast aur adiabatic hai — koi heat leak nahi hoti) lekin yeh irreversible hai: yeh gas ko stir up karta hai. Irreversibility entropy $s$ se measure hoti hai, aur shock ke across $s_{2} > s_{1}$ . Us entropy jump ki bookkeeping cost stagnation pressure $p_{0}$ mein ek drop hai (woh pressure jo flow tab reach karta jab smoothly rest tak laaya jaata):

$\frac{p _{0, 2}}{p _{0, 1}} = exp (always \leq 0 - Δ s / R), Δ s = s_{2} - s_{1} \geq 0.$

$p_{0, 1}, p_{0, 2}$ = stagnation pressure shock se pehle / baad.
$Δ s$ = shock ke across entropy rise (kabhi negative nahi — yahi second law hai).
$R$ = gas constant, sirf units fix karta hai.
Ek negative number ka exponential 1 se kam hota hai ⇒ stagnation pressure hamesha drop karti hai.

WOH PEECHE KI FORCE KE BARABAR KYUN HAI. Stagnation pressure flow ka "push potential" hai. Ek wake jisme oncoming stream se kam $p_{0}$ ho, woh momentum deficit carry karta hai. Body ke around ek box par momentum balance karo aur woh deficit ek net body par peeche ki force ke roop mein show hoti hai — drag. Koi stickiness (viscosity) nahi chahiye; shock ne akele yeh kar diya. Dekho Entropy and stagnation pressure loss.

PICTURE. Ek control box: aage full stagnation pressure, andar ek shock, wake mein reduced $p_{0}$ , body par reaction drag dikhata arrow.

Step 7 — Transonic spike, aur kyun supersonic drag phir se girta hai

KYA. $M_{\infty} = 1$ se neeche bhi, curved wing top par air speed up hoti hai aur locally $M = 1$ hit kar sakti hai. Sabse kam $M_{\infty}$ jahan yeh pehli baar hota hai woh critical Mach number $M_{cr}$ hai. Thoda aur tez $M_{dd}$ tak push karo aur ek supersonic pocket form hoti hai, jiske upar ek shock hai jo (Step 6) drag aur uske peeche boundary layer mota karta hai. Tab drag spike ho jaati hai — $M_{\infty} \approx 0.8$ – $1.0$ ke paas drag-divergence rise.

BAAD MEIN KYUN GIRTA HAI. $M_{\infty} = 1$ past karo har jagah aur shocks oblique ho jaate hain (Step 5) — har unit length per weak, body ke along peeche ki taraf lean karte hue. Neeche ka wave-drag recipe ek factor $1/ M_{\infty}^{2} - 1$ carry karta hai jo $M_{\infty}$ badhne par shrink karta hai. Toh curve 1 ke paas hump up hoti hai aur phir slide back down karta hai.

PICTURE. Ek drag-coefficient-vs- $M_{\infty}$ curve: flat, phir $M_{cr} \to M_{dd} \to 1$ par ek sharp hill, phir gentle decline.

Formula Supersonic pay-off (Ackeret thin-airfoil)

$c_{d, wave} = \frac{4}{M _{\infty}^{2} - 1} (α^{2} + \overline{(d y_{c} / d x)^{2}} + \overline{(d y_{t} / d x)^{2}})$

$\frac{4}{M _{\infty}^{2} - 1}$ = woh " $1/ M$ -shrinks-it" factor. $M_{\infty}^{2} - 1$ quantity Prandtl–Glauert compressibility factor hai (bahut si books mein $β$ likha jaata hai — ek alag $β$ Step 5 ke shock angle se; same letter, unrelated meaning). Dekho Prandtl-Glauert compressibility correction.
$α$ = angle of attack = airfoil ki chord line ka oncoming flow ke relative tilt; $α^{2}$ = lift banane ke liye tilting se drag (har term ek square hai ⇒ drag kabhi negative nahi).
$\overline{(d y_{c} / d x)^{2}}$ = camber-line slope ka chord-averaged square ( $y_{c}$ = mean line, toh $d y_{c} / d x$ = mean line kitna rise karta hai); overbar ka matlab hai "chord ke along average."
$\overline{(d y_{t} / d x)^{2}}$ = thickness slope ka chord-averaged square ( $y_{t}$ = half-thickness); zero lift par bhi present hai ⇒ wings thin banao.

Ek-picture summary

Ek ring → bahut saari rings → ek Mach cone (angle $μ$ , $sin μ = 1/ M$ ) → real shock steeper lean karta hai ( $β > μ$ ) → uske across entropy badhti hai → stagnation pressure girta hai → woh deficit drag hai.

Recall Feynman retelling — plain words mein bolo

Ek flying cheez air ko tap karti rehti hai, aur har tap speed of sound par ek pressure ring ki tarah spread hoti hai. Agar cheez slow hai, uski rings aage nikal jaati hain aur air politely side mein ho jaati hai — koi problem nahi. Agar cheez sound se tez hai, toh woh apni rings ko outrun kar leti hai: woh aage ki air ko warn nahi kar sakti, toh saari rings ek slanted line ke neeche bunch up ho jaati hain, Mach cone, jiska steepness $sin μ = 1/ M$ sirf "ring radius over how far the thing moved" hai. Ab aage ki air body ke baare mein sirf last instant mein jaanti hai aur side mein ease karne ki jagah nahi hai, toh woh apna direction change ek razor-thin shock mein slam kar deti hai. Ek real shock ek genuine squeeze hai, toh woh silent Mach cone se zyada steeply lean karta hai. Air ko roughly squeeze karna use stir up karta hai — entropy badhti hai — aur woh stirring chupke se flow ka push chura leti hai, uska stagnation pressure gira deti hai. Wake mein aage se kam push ka matlab hai air body ko peeche dhakka de rahi hai: woh peeche ka push, jo sirf shocks se paida hua aur jisme bilkul bhi friction ki zaroorat nahi thi, wave drag hai. Yeh sound ki speed ke aas paas screams upward (pehle wing par ek shock sprout hota hai) aur phir ease off hota hai jab tum fully supersonic ho aur shocks weaker slanted lines mein peeche lean ho jaate hain.

Recall Quick checks

$sin μ = 1/ M$ geometrically kahan se aata hai? ::: Ek right triangle jisme ring radius $a t$ $μ$ ke opposite hai aur body travel $u t$ hypotenuse hai; sine = opposite/hypotenuse = $a / u = 1/ M$ . Ek real shock Mach cone se steeper kyun hota hai? ::: Mach cone zero-strength limit hai ( $θ \to 0$ ); koi bhi finite turn finite compression chahta hai, jo $β > μ$ force karta hai. Wave drag physically kahan se aata hai? ::: Shock ke across entropy badhti hai ⇒ stagnation pressure girta hai ⇒ wake momentum deficit ⇒ net rearward force, bina kisi viscosity ke. Supersonic wave drag Mach ke saath kyun decrease hota hai? ::: Kyunki Ackeret formula mein har term $1/ M_{\infty}^{2} - 1$ se multiply hota hai, aur woh factor $M_{\infty}$ badhne par chhota hota jaata hai; physically shocks oblique aur weaker ho jaate hain har unit length par.