Oblique shock waves — θ-β-M relation: Visual walkthrough

3.1.13 · D2 · Physics › Compressible Flow & Aerodynamics › Oblique shock waves — θ-β-M relation

Step 1 — Koi bhi symbol aane se pehle ki picture: ek fast flow ek corner se milti hai

KYA. Hawa ka ek stream tezi se (sound se bhi tezi) ek wall ke saath chal raha hai. Wall achanak flow ke andar ki taraf ek chhote angle se mod leti hai. Hawa ek sharp inward corner ke around smoothly flow nahi kar sakti, isliye hawa mein ek patli, seedhi, tilted line ban jaati hai — ek shock. Flow us line ko cross karti hai, mod leti hai, aur slow ho jaati hai.

YUN yahan se kyun shuru karein. Neeche sab kuch ek drawing par bookkeeping hai. Agar aap yeh drawing apne dimaag mein rakho, koi bhi formula surprising nahi lagega. Hum bas teen angles ka naam rakh rahe hain.

PICTURE. Figure dekho.

Left se aata cyan arrow incoming velocity hai. Iska length $V_{1}$ rakh lo (speed 1 = "upstream").
Amber line shock hai. Yeh incoming arrow ke saath jo angle banati hai woh $β$ hai (the "shock-wave angle"). Hum $β$ incoming flow se shock tak measure karte hain.
Cross karne ke baad, flow nayi wall follow karta hai. Purani direction aur nayi direction ke beech ka angle $θ$ hai (the "deflection angle" — flow kitna mod gaya).

Figure — Oblique shock waves — θ-β-M relation

Yeh bilkul wahi setup hai jo Supersonic Wedge and Cone Flow mein hai: wall bend hi wedge half-angle hai.

Step 2 — Ek clever move: har velocity ko "shock mein" aur "shock ke saath" mein split karo

KYA. Incoming arrow lo aur usse shock line ko reference banake do pieces mein todo:

ek piece shock ke perpendicular (seedha uske andar pointing), aur
ek piece shock ke parallel (uske saath saath sliding).

Yehi kaam outgoing arrow ke saath bhi karte hain.

YUN yeh tool kyun, koi aur kyun nahi? Ek shock achanak pressure change ki ek wall hai. Pressure kisi surface ke perpendicular push karta hai. Toh shock flow ko sirf seedha apne aap ke across hi shove kar sakti hai. Along-the-sliding direction ko koi push nahi milta. Perpendicular + parallel mein split karna isliye natural coordinate system hai — yeh "woh direction jo change hoti hai" ko "woh direction jo nahi hoti" se alag kar deta hai. Koi bhi doosra axis dono ko tangle kar deta.

PICTURE. Right-angle boxes split dikhate hain.

$u_{n 1} = V_{1} sin β$ — incoming arrow ka normal (perpendicular) part. Yahan $sin β$ isliye aaya kyunki hypotenuse $V_{1}$ aur corner par angle $β$ wale right triangle ki perpendicular leg hai (hypotenuse) $\times sin β$ .
$w_{1} = V_{1} cos β$ — tangential (along-shock) part. Adjacent leg hai (hypotenuse) $\times cos β$ .

Subscript convention: $n$ = normal, $t$ ya $w$ = tangential, $1$ = upstream, $2$ = downstream.

Step 3 — Tangential slide untouched rehta hai (yeh master fact hai)

KYA. Along-the-shock speed pehle aur baad mein same rehti hai: $w_{1} = w_{2} ⟹ V_{1} cos β = V_{2} cos (β - θ) .$

KYU. Shock ke saath koi force act nahi karta (Step 2 ki wajah: pressure sirf us par perpendicular push karta hai). Koi sideways force nahi ⇒ sideways momentum mein koi change nahi ⇒ tangential speed conserved hai. Yeh ek sentence hi oblique shocks ke peeche ka poora trick hai.

YUN $(β - θ)$ kyun aata hai. Outgoing flow $θ$ se wall ki taraf bend ho chuki hai. Toh outgoing arrow aur shock line ke beech ka angle ab $β$ nahi raha — woh $β$ minus bend $θ$ hai. Figure dekho: downstream triangle usi amber shock line ke saath angle $(β - θ)$ par baitha hai.

Step 4 — Do velocity facts ko angles ke ek ratio mein badlo

KYA. Do normal-component equations ko divide karo, phir tangential fact se $V_{2} / V_{1}$ khatam karo.

Normal speeds ke ratio se shuru karo: $\frac{u _{n 2}}{u _{n 1}} = \frac{V _{2} s i n ( β - θ )}{V _{1} s i n β} .$

Step 3 se, $\frac{V _{2}}{V _{1}} = \frac{cos β}{cos ( β - θ )}$ . Substitute karo: $\frac{u _{n 2}}{u _{n 1}} = \frac{c o s β}{c o s ( β - θ )} \cdot \frac{s i n ( β - θ )}{s i n β} = \frac{t a n ( β - θ )}{t a n β} .$

YUN yeh kyun karo. Hum ek aisi equation chahte hain jisme koi speed na bache — sirf angles aur Mach number hon. Speeds $V_{1}, V_{2}$ unknown aur annoying hain; ratio trick unhe cancel kar deta hai. Jo bachta hai woh ek pure-geometry statement hai: shock ke across fractional slow-down tangents ke ratio ke barabar hai.

YUN tangent specifically kyun? $tan = \frac{sin}{cos} = \frac{opposite}{adjacent} = \frac{across-shock part}{along-shock part}$ . Tangent exactly woh ratio hai jinhe hum split kiya do legs ka. Toh $tan β$ aur $tan (β - θ)$ literally measure karte hain "flow shock mein kitni steeply dive karta hai" pehle aur baad mein.

PICTURE. Figure upstream triangle aur downstream triangle ko stack karta hai jo same base $w$ share karte hain (kyunki $w$ conserved hai). Heights $u_{n 1}$ aur $u_{n 2}$ hain. Same base, downstream shorter height ⇒ flow shock ke against flatten ho gayi ⇒ $tan (β - θ) < tan β$ .

Step 5 — Physics feed karo: andar chhupa hua normal shock

KYA. Ab tak sirf triangles the. Ab ek physics law use karo. Perpendicular direction ke saath, flow bilkul ek normal shock ki tarah behave karti hai Mach number ke saath: $M_{n 1} = M_{1} sin β .$ Normal shock ka mass-conservation + Rankine–Hugoniot result velocity drop deta hai: $\frac{u _{n 2}}{u _{n 1}} = \frac{ρ _{1}}{ρ _{2}} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{( γ + 1 ) M _{n 1}^{2}} .$

YUN hum normal-shock law borrow kar sakte hain. Step 2 mein split karne ka poora point yahi tha: perpendicular slice ek self-contained normal shock hai, jo ek unchanged sideways slide par ride kar raha hai. Toh normal-shock ka har formula apply hoga agar aap usse $M_{n 1}$ dein, $M_{1}$ nahi.

Symbols padhna.

$M_{n 1} = M_{1} sin β$ ::: normal problem ke liye effective upstream Mach. Sirf yahi part "shock dekhta hai".
$γ$ ::: ratio of specific heats, hawa ke liye $1.4$ — yeh batata hai gas kitni squishy hai.
Right-hand fraction ::: hamesha $< 1$ hoti hai jab $M_{n 1} > 1$ (flow slow hoti hai), aur $= 1$ jab $M_{n 1} = 1$ (koi shock nahi).

PICTURE. Figure perpendicular strip ko isolate karta hai aur use ek mini normal-shock ki tarah dikhata hai: incoming $u_{n 1}$ , ek thin shock, slower $u_{n 2}$ , sideways slide $w$ faded draw karke remind karta hai ki woh sirf saath chal rahi hai.

Step 6 — Do ratios ko equate karo aur θ-β-M relation padho

KYA. Step 4 (geometry) aur Step 5 (physics) dono usi ratio $u_{n 2} / u_{n 1}$ ke do expressions hain. Unhe equal karo: $\frac{t a n ( β - θ )}{t a n β} = \frac{( γ - 1 ) M _{1}^{2} s i n ^{2} β + 2}{( γ + 1 ) M _{1}^{2} s i n ^{2} β} .$

KYU. Ek ratio khud ke barabar hota hai. Geometry ne ek side banaayi, physics ne doosri; unhe agree karne par force karna exactly woh condition hai ki kin angles ke liye ek real oblique shock exist karta hai.

Clean up karna. $tan (β - θ)$ ko expand karke aur $tan θ$ ke liye solve karke (routine trig algebra, neeche verify ki gayi) sab kuch compact form mein aa jaata hai:

Figure ek fixed $M_{1}$ ke liye $θ$ ko $β$ ke against plot karta hai: woh famous hump. Use left se right trace karo aur agli step mein naam diye gaye do special crossings padho.

Step 7 — Har edge case, seedha numerator aur hump se padho

Single term $(M_{1}^{2} sin^{2} β - 1)$ saare degenerate cases control karta hai. Iska sign dekho.

Case A — $sin β = 1/ M_{1}$ (sabse weak shock, ek Mach wave). Yahan $M_{1}^{2} sin^{2} β - 1 = 0$ , toh $tan θ = 0$ ⇒ $θ = 0$ . Koi bending nahi. Shock infinitely weak hai: yeh Mach angle $μ = arcsin (1/ M_{1})$ par ek Mach wave hai. Yeh hump ka leftmost foot hai.

Case B — $β = 9 0^{\circ}$ (straight-on). $cot 9 0^{\circ} = 0$ , toh $tan θ = 0$ ⇒ $θ = 0$ dobara. Flow ke perpendicular shock kuch nahi modti — yeh ek pure normal shock hai. Yeh hump ka rightmost foot hai.

Case C — peak $θ_{m a x}$ . In do feet ke beech curve ek maximum $θ_{m a x}$ tak rise karta hai aur wapas neeche aata hai. Kisi bhi $θ < θ_{m a x}$ ke liye horizontal line hump ko do baar kaat ti hai:

weak solution (chhota $β$ , peak ke left) — usually wahi jo nature choose karta hai, downstream often abhi bhi supersonic;
strong solution (bada $β$ , peak ke right) — downstream subsonic.

Case D — $θ > θ_{m a x}$ . Line hump miss kar jaati hai: koi real $β$ exist nahi karta. Attached shock form nahi ho sakta; woh body se ek curved bow shock ki tarah pop off ho jaata hai.

Case E — $sin β < 1/ M_{1}$ (numerator negative). $tan θ$ negative aata — physically impossible "expansion turn." Woh regime Prandtl-Meyer Expansion ka hai (ek convex corner), shock ka nahi. Isliye hum simply $β$ ko Mach angle se neeche forbid karte hain.

Ek-picture summary

Yeh final figure poori chain compress karta hai: ek incoming arrow → normal + tangential mein split → tangential conserved, normal ek normal shock karta hai → do velocity ratios equate karo → θ-β-M curve apne Mach-wave foot, apne hump peak $θ_{m a x}$ , aur apne normal-shock foot ke saath. Arrows ko s02 → s07 order mein trace karo.

Recall Poore walkthrough ki Feynman retelling

Hawa ki ek fast river ek tilted line (shock) se takraati hai. Main river ki speed ko do mein split karta hoon: woh part jo seedha line mein ja raha hai, aur woh part jo uske saath saath slide kar raha hai. Sliding part shock ki parwah nahi karta — kuch sideways push nahi karta — toh woh dono sides par bilkul same rehta hai. Seedha-andar wala part slam ho jaata hai, aur woh wahi rules follow karta hai jo ek head-on (normal) shock mein hote hain, lekin sirf tab agar main usse uski apni chhoti speed $M_{1} sin β$ doon. Ab mere paas yeh likhne ke do tarike hain ki seedha-andar wala part kitna slow hua: ek pure triangle geometry se ( $tan (β - θ) / tan β$ ), ek shock physics se (density formula). Woh same slow-down describe karte hain, toh main unhe equal karta hoon. Rearrange karne se ek tidy equation milti hai jo turn angle $θ$ , shock tilt $β$ , aur speed $M_{1}$ ko link karti hai. Iska numerator $M_{1}^{2} sin^{2} β - 1$ padh ke poori kahani milti hai: gentle Mach wave aur head-on shock par zero (dono taraf koi bending nahi), beech mein ek hump jiska peak woh sabse bada turn hai jo ek clean shock kar sakta hai — isse zyada maango aur shock give up karke body se curve off ho jaati hai.