Shock wave angle, deflection angle: Visual walkthrough

3.1.14 · D2 · Physics › Compressible Flow & Aerodynamics › Shock wave angle, deflection angle

Step 1 — Ek supersonic flow "kya dekhta hai" jab wo wedge se milta hai

KYA. Imagine karo air seedha right side mein stream kar rahi hai steady speed par, sound se tez. Wo ek solid wedge se takraati hai — ek triangle jo hawa mein point kar raha hai. Wedge ki surface kuch angle se upar tilt hoti hai. Air ko raasta banana padta hai.

KYUN. Aam (slow) flow mein, air ko wall aate hue feel hota hai aur wo smoothly uske around curve karti hai — pressure "messages" upstream travel karti hain aur use warn karti hain. Lekin ye messages speed of sound par travel karti hain. Agar air sound se tez move kar rahi hai, koi bhi message use outrun nahi kar sakta. Toh air ko koi warning nahi milti: wo ek razor-thin sheet se takraati hai aur ek saath turn karti hai. Woh sheet hi shock hai.

PICTURE. Do straight lines matter karti hain. Ek khud shock hai, wedge tip se peeche jhukti hui. Doosri wedge surface hai. Do angles yahan janam lete hain — aur unhe alag rakhna hi poori kahani hai.

Figure — Shock wave angle, deflection angle

Step 2 — Velocity ko do honest arrows mein split karo

KYA. Incoming velocity lo — ek arrow jiska length $V_{1}$ hai aur seedha hawa mein point kar raha hai. Hum ise do arrows se replace karte hain jo milke isse add up karte hain: ek shock face mein seedha pointing (shock ke perpendicular), ek shock face ke along pointing.

KYUN. Shock ek pressure ki wall hai. Pressure hamesha surface ke perpendicular push karta hai (jaise paani dam ke flat against press karta hai). Toh shock flow ko sirf apne aap se across shove kar sakta hai — apne aap ke along uski koi grip nahi (hum assume karte hain no friction, i.e. inviscid flow). Matlab:

velocity ka face-ke-along part untouched guzar jaata hai;
sirf face-mein-jaane-wala part slam hota hai, squash hota hai, aur slow hota hai — bilkul ek plain normal shock ki tarah.

Yeh ek insight hi poori mushkil problem ko normal-shock problem mein convert kar deti hai.

PICTURE. $V_{1}$ aur uske do pieces se bana right triangle. Yahan $sin β$ = opposite/hypotenuse perpendicular piece nikaalti hai, aur $cos β$ = adjacent/hypotenuse along piece nikaalti hai.

Step 3 — Downstream: same do pieces, lekin turn ke saath

KYA. Shock ke baad, flow ki nayi speed $V_{2}$ aur nayi direction hai — $θ$ se bent. Hum is arrow ko bhi same tarah split karte hain: ek perpendicular piece $u_{2}$ aur along-piece $w_{2}$ .

KYUN. Hume "before" aur "after" ko same axes par compare karna hai (perpendicular vs along the shock). Kyunki outgoing flow shock face ke saath angle $(β - θ)$ banati hai — $β$ nahi — toh split $(β - θ)$ use karta hai.

$(β - θ)$ kahan se aata hai? Shock incoming flow se $β$ par jhukta hai. Outgoing flow $θ$ wall ki taraf closer hai (wo wall ki taraf mudi). Toh outgoing flow aur shock face ke beech bacha angle $β - θ$ hai. Ise seedha figure se padho.

PICTURE. Downstream right triangle, shock ke saath $(β - θ)$ angle par, apni perpendicular aur along legs ke saath labelled.

Step 4 — Conservation of mass se velocity ratio milta hai

KYA. Mass thin shock ke andar pile nahi ho sakta ya vanish nahi ho sakta. Jo bhi per second face mein flow karta hai woh bahar nikalta hai. Symbols mein, $ρ_{1} u_{1} = ρ_{2} u_{2}$ , jahan $ρ$ (Greek "rho") density hai — kitna mass har cubic metre mein hota hai.

KYUN sirf normal piece? Mass shock ko uske through move karke cross karta hai — woh perpendicular motion hai. Along-the-face gliding kuch bhi across carry nahi karti. Toh continuity $u$ use karta hai, $w$ nahi.

PICTURE. Shock ko straddling karta ek chhota box: left face mein same mass rate jitna right face se bahar.

Perpendicular speeds ka ratio paane ke liye rearrange karo: $\frac{u _{2}}{u _{1}} = \frac{ρ _{1}}{ρ _{2}}$

Left side: perpendicular speed kitni drop hui.
Right side: flow kitna compress hua uska inverse. Peeche denser ⇒ peeche perpendicular flow slower. Yahi toh "traffic bunches up jab wo slow hoti hai" hai.

Step 5 — Normal-shock density law borrow karo

KYA. Ek normal shock ke liye, Rankine–Hugoniot density jump sirf normal Mach number $M_{n 1}$ par depend karta hai: $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 ) M _{n 1}^{2}}{( γ - 1 ) M _{n 1}^{2} + 2}, M_{n 1} = M_{1} sin β$

KYUN $M_{n 1} = M_{1} sin β$ ? Mach number hai speed ÷ sound-speed. Speed ka sirf perpendicular piece shock feel karta hai, aur woh piece $V_{1} sin β$ hai. Same sound speed se divide karo aur $M_{1} sin β$ milta hai. Hum $M_{n 1}$ use karte hain — $M_{1}$ nahi — kyunki normal shock us flow ko react karta hai jo use head-on hit karta hai, aur sirf perpendicular part head-on hit karta hai.

KYUN yeh tool? Hume density aur flow ke beech ek known relationship chahiye thi. Normal-shock law woh known cheez hai — aur hum ise use karne ka haq earn kar chuke hain kyunki Step 2 ne prove kiya ki perpendicular flow ek normal shock hai.

PICTURE. Disguise reveal hua: apni aankhen rotate karo taaki shock vertical ho, aur perpendicular flow ek textbook normal shock hai; along-flow sirf ek constant sideways drift hai.

Step 4 ke saath combine karo: $\frac{u _{2}}{u _{1}} = \frac{ρ _{1}}{ρ _{2}} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{( γ + 1 ) M _{n 1}^{2}}$ (Humne fraction flip kiya kyunki hume $ρ_{1} / ρ_{2}$ chahiye, reciprocal.)

Step 6 — Speeds ko $tan$ se angles mein convert karo

KYA. Har side par, perpendicular piece ko along piece se divide karo: $tan β = \frac{u _{1}}{w _{1}}, tan (β - θ) = \frac{u _{2}}{w _{2}} = \frac{u _{2}}{w _{1}}$

KYUN $tan$ , aur kyun abhi? Ek right triangle par, $tan$ (angle) = opposite ÷ adjacent = perpendicular piece ÷ along piece. Yeh exactly hamare do velocity components ka ratio hai. Toh $tan$ woh tool hai jo velocity ratio ko us angle mein convert karta hai jis ki humhe actually care hai. Hum ise abhi use karte hain kyunki finally dono sides par dono components hain.

Key cancellation. Downstream side par $w_{2} = w_{1}$ (Step 2). Toh jab hum do tangents divide karte hain, along-piece $w_{1}$ cancel ho jaati hai aur sirf perpendicular ratio bachta hai: $\frac{t a n ( β - θ )}{t a n β} = \frac{u _{2} / w _{1}}{u _{1} / w _{1}} = \frac{u _{2}}{u _{1}}$

PICTURE. Dono triangles ek hi horizontal base $w_{1}$ share karte hain; sirf unki heights ( $u_{1}$ vs $u_{2}$ ) alag hain. Woh shared base hi wajah hai ki ratio clean up hota hai.

Ab Steps 4–6 ko stitich karo: $\frac{tan ( β - θ )}{tan β} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{( γ + 1 ) M _{n 1}^{2}}$ Yahi θ–β–M relation hai — bas abhi tak apni sabse pretty form mein comb nahi hua.

Step 7 — Ise standard compact form mein comb karo

KYA. $M_{n 1} = M_{1} sin β$ substitute karo, tangents ko $tan = sin / cos$ se expand karo, aur identity $cos 2 β = 1 - 2 sin^{2} β$ use karo. Algebra ki dhool settle hone ke baad:

KYUN yeh form? Yeh $tan θ$ ko akele left side par isolate karta hai, toh tum $β$ ko small se large tak sweep kar sakte ho aur $θ$ plot kar sakte ho — jisse hum next weak/strong branches discover karte hain.

PICTURE. Fixed $M_{1}$ ke liye $θ$ -vs- $β$ curve: zero se rise karti, maximum par crest karti, wapas zero par fall karti.

Step 8 — Har case, curve se padho

KYA & KYUN. Step 7 ki curve batati hai ki ek given wedge kitne shocks produce karta hai. Iske across chalo:

PICTURE. Wahi curve, ab tumhare chosen $θ$ par ek horizontal line ke saath: yeh curve ko do baar cross karti hai (weak + strong), ek baar graze karti hai ( $θ_{m a x}$ par), ya bilkul miss karti hai (detached).

The one-picture summary

Sab kuch ek canvas par: incoming arrow perpendicular ( $u_{1}$ ) aur along ( $w_{1}$ ) mein split; shock ek leaning line ke roop mein $β$ par; outgoing arrow $θ$ se bent, apni $u_{2}$ (chhoti) aur $w_{2} = w_{1}$ (unchanged) mein split; aur yaad dilana ki perpendicular story ek normal shock hai.

Recall Feynman retelling — plain words mein poora walkthrough

Air ek wedge mein itni tez race karti hai ki warn nahi ho sakti, toh wo ek thin leaning line par crash karti hai — shock, angle $β$ par tilted. Hum air ki speed ko do arrows mein split karte hain: ek us line mein seedha poking, ek us par sliding. Sliding wala shock ki parwah nahi karta aur unchanged guzar jaata hai — kyunki shock ka push sab taraf sideways hai, seedha apni face mein, along uske kuch nahi grip karta. Sirf poking-in arrow crush hoti hai aur slow hoti hai, bilkul ek plain head-on (normal) shock ki tarah, jiske rules hum pehle se jaante hain. "Kitna slow hua?" wahi hai "kitna squish hua?" (traffic bunches jab slow hoti hai) — aur squish law sirf poking-in Mach number, $M_{1} sin β$ , chahiye hai. Finally, har side par poke ko slide se divide karna speeds ko $tan$ ke through angles mein convert karta hai, sliding arrow cancel ho jaata hai kyunki dono sides par same hai, aur θ–β–M formula nikal aata hai. Ise plot karo aur poori life story dikhti hai: ek gentle Mach wave ek end par, doosre par ek square normal shock, do shocks (weak aur strong) kisi bhi modest turn ke liye beech mein, aur — agar tum bahut sharp turn demand karo — koi attached shock nahi, bas ek bow shock aage float karta hua.

#recall

Normal velocity component hi shock se kyun change hoti hai?

Pressure shock face ke perpendicular act karta hai aur koi friction nahi hai, toh face ke along koi force act nahi karta — tangential (along) component conserved rehta hai, sirf normal wala compress hota hai.

M_{n 1} = M_{1} sin β

kyun hai aur

M_{1} cos β

kyun nahi?

Perpendicular velocity

V_{1} sin β

hai (velocity triangle par

β

ke opposite wali side); sound speed se divide karo toh perpendicular Mach number milta hai.

Step 6 mein along-component

w_{1}

cancel kyun ho jaata hai?

Kyunki

w_{2} = w_{1}

, toh jab tum

tan (β - θ) / tan β = (u_{2} / w_{1}) / (u_{1} / w_{1})

divide karte ho toh common

w_{1}

cancel ho jaata hai,

u_{2} / u_{1}

bachta hai.

β = 9 0^{\circ}

par

θ = 0

kyun hota hai?

cot 9 0^{\circ} = 0

, toh

tan θ = 0

— ek normal shock flow ko kuch nahi turn karta.

Formula mein

M_{1}^{2} sin^{2} β - 1

kya represent karta hai?

Shock strength

M_{n 1}^{2} - 1

; positive = real shock, zero = Mach wave.

θ > θ_{m a x}

ke liye kya banta hai?

Body ke aage ek detached curved bow shock — koi attached oblique shock exist nahi karta.