Shock wave angle, deflection angle: Question bank — Hinglish deep dive

3.1.14 · D5 · Physics › Compressible Flow & Aerodynamics › Shock wave angle, deflection angle

Traps se pehle, har symbol ko page par ek picture ke saath rakh lete hain, taaki koi bhi cheez bina samjhe use na ho.

Sketch dekho: incoming arrow, tilted shock line, aur bent outgoing arrow — bas itna hi hum kabhi draw karte hain. Figure s01 dikhata hai ki ek single velocity arrow kaise shock-ke-andar-jaane-wale piece aur shock-ke-saath-chalne-wale piece mein split hota hai — yahi poora trick hai.

Figure — Shock wave angle, deflection angle

Ab geometry step-by-step: sirf normal piece squash hota hai (jaise ek normal shock), jabki tangential piece unchanged guzar jaata hai. Figure s02 dikhata hai yeh reduction aur kahan $M_{n 1}$ , $M_{n 2}$ , aur turn $θ$ rehte hain.

Aakhir mein, core relation jis par poora bank tika hai — parent $\theta$–$\beta$–$M$ relation:

Yeh page apne neighbours Mach Angle and Mach Waves, Rankine-Hugoniot Relations, Prandtl-Meyer Expansion, Detached Bow Shock, aur Supersonic Wedge & Cone Flow par bhi build karta hai.

True or false — justify karo

A normal shock, oblique shock ka ek special case hai.

True — yeh

β = 9 0^{\circ}

wala oblique shock hai, jahan poori velocity normal hoti hai aur kuch bhi tangential nahi, isliye

θ \to 0

Wedge half-angle

θ

badhane par weak branch mein wave angle

β

hamesha badhta hai.

Weak branch par

θ_{m a x}

tak True hai — zyaada turning ke liye steeper shock chahiye; lekin

θ_{m a x}

ke baad koi attached

β

nahi hota (shock detach ho jaata hai).

Ek fixed

M_{1}

ke liye, har deflection angle

θ < θ_{m a x}

exactly ek shock angle

β

se correspond karta hai.

False — do hote hain: ek weak shock (chhota

β

, flow supersonic rehti hai) aur ek strong shock (bada

β

, flow subsonic ho jaati hai).

Tangential velocity oblique shock ke across conserve hoti hai kyunki shock ke saath-saath flow slow hoti hai.

False — yeh isliye conserve hoti hai kyunki (inviscid) shock pressure sirf apne face ke perpendicular direction mein exert karta hai, isliye tangential force zero hoti hai chahe tangential flow kitni bhi tez ho.

Fixed

θ

par

M_{1}

badhane se weak-shock wave angle

β

kam hota hai.

True — ek faster flow utni hi turning ek zyaada oblique (shallower) shock ke through kar sakti hai, isliye

β

(bhi-ghatte hue) Mach angle ki taraf girta hai.

Maximum deflection angle

θ_{m a x}

har Mach number ke liye same hota hai.

False —

θ_{m a x}

M_{1}

ke saath badhta hai; ek faster stream detach hone se pehle ek bade attached-shock angle tak turn ho sakti hai.

Mach angle limit par (

β = μ

), "shock" koi compression carry nahi karta.

True — wahan

M_{n 1} = M_{1} sin μ = 1

, isliye normal Mach number exactly 1 hai: ek infinitely weak wave (Mach wave), koi pressure jump nahi,

θ = 0

Ek weak oblique shock ke peechhe flow ab bhi supersonic ho sakti hai chahe

M_{n 2} < 1

ho.

True — normal component subsonic hai (

M_{n 2} < 1

), lekin

M_{2} = M_{n 2} / sin (β - θ)

ek chhote sine se divide hota hai, jo total speed ko supersonic restore kar deta hai.

Dono lower limit (

β = μ

) aur upper limit (

β = 9 0^{\circ}

) zero deflection dete hain.

True —

μ

par shock vanishingly weak hai;

9 0^{\circ}

par yeh ek normal shock hai purely head-on flow ke saath, isliye dono cases mein streamline bend nahi karti.

Error dhundo

"Kyunki algebra normal shock se match karta hai, density-ratio formula mein directly

M_{1}

plug karo."

Error: tumhe har normal-shock relation mein

M_{1}

ki jagah normal Mach number

M_{n 1} = M_{1} sin β

use karna hoga — sirf perpendicular component shock hota hai.

"Deflection angle, corner par wave angle ke barabar hota hai."

Error:

β

incoming flow se shock line tak measure kiya jaata hai,

θ

streamline ka bend hai; hamesha

θ < β

hota hai, aur

β = 9 0^{\circ}

par hame

θ = 0

milta hai — yeh genuinely alag hain.

"Downstream Mach number recover karne ke liye

M_{2} = M_{n 2}

use karo."

Error:

M_{n 2}

sirf shock ke normal component hai; full speed

M_{2} = M_{n 2} / sin (β - θ)

hai kyunki tangential component downstream bhi present rehta hai.

"Oblique-shock relation tumhe

θ

aur

M_{1}

se uniquely

β

deta hai."

Error:

θ

–

β

–

M

relation ko invert karne par do

β

roots milte hain (weak aur strong); ek teesra "solution"

β

near

μ

par sirf tab Mach wave mein collapse hota hai jab

θ = 0

ho.

"Ek bada wedge bas ek bada, steeper, hamesha-attached shock banata hai."

Error: jab

θ > θ_{m a x}

ho jaata hai, koi attached oblique shock exist nahi karta — shock body ke aage ek curved bow shock mein detach ho jaata hai (Detached Bow Shock).

"Shock tangential velocity ko bhi slow karta hai, bas normal part se kam."

Error: tangential velocity bilkul unchanged rehti hai (

w_{2} = w_{1}

); zero tangential force ka matlab zero tangential deceleration — yeh derivation ki cornerstone hai.

"Agar

M_{1} sin β < 1

hai toh bhi hume compression shock milta hai."

Error: shock exist karne ke liye

M_{n 1} = M_{1} sin β

ko 1 se zyaada hona chahiye; agar yeh 1 se neeche hai toh koi shock form nahi hota (tumne jo

β

choose kiya wo Mach angle se shallower hai, jo shock ke liye physically impossible hai).

Why questions

Hum velocity ko normal aur tangential components mein split kyun karte hain?

Kyunki sirf normal component normal shock ki tarah compress hota hai jabki tangential unchanged guzarta hai — split (Figure s01) ek 2-D oblique problem ko ek 1-D normal-shock problem plus geometry mein reduce kar deta hai.

M_{n 1} = M_{1} sin β

(sine, cosine nahi) kyun hai?

β

flow se shock tak measure kiya jaata hai, isliye shock face mein jaane wala component

V_{1} sin β

hai; sound speed se divide karne par

M_{1} sin β

milta hai.

θ

-vs-

β

curve dono ends par zero par kyun return karta hai?

β = μ

par shock infinitely weak hai (koi turning nahi), aur

β = 9 0^{\circ}

par flow head-on aata hai koi sideways deflection ke bina — 0 se uthne aur 0 par return karne wali curve ko beech mein peak karna hi padega, jo

θ_{m a x}

deta hai (Figure s03).

θ_{m a x}

exceed karne par attached shock ki jagah detached shock kyun force hota hai?

Koi bhi attached geometry flow ko

θ_{m a x}

se zyaada turn nahi kar sakti; flow wedge par abruptly turning se "haath utha leta hai" aur ek curved bow shock ke through compress ho jaata hai jo aage khada hota hai, jahan local turning har jagah limits ke andar rehti hai.

External aerodynamics mein weak solution hi zyaadatr kyun dikhai deta hai?

Weak shock flow ko supersonic rehne deta hai aur free flight ke low-back-pressure environment se match karta hai; strong (subsonic) branch ko typically ek downstream pressure constraint chahiye jo external flow impose nahi karta.

Shock face ke across mass conservation sirf normal velocity kyun use karta hai?

Mass shock ke through sirf face mein move karke cross karta hai, yaani normal component ke through;

ρ_{1} u_{1} = ρ_{2} u_{2}

, isliye

ρ_{2} / ρ_{1} = u_{1} / u_{2}

Mach angle

μ = sin^{- 1} (1/ M_{1})

sabse chhota possible

β

kyun hai?

Ek shock ko

M_{n 1} = M_{1} sin β \geq 1

chahiye; ise satisfy karne wala sabse chhota

β

sin β = 1/ M_{1}

se milta hai, yaani

β = μ

— koi bhi shallower aur normal Mach 1 se neeche chala jaayega.

Edge cases

Jab

β \to 9 0^{\circ}

ho toh

θ

aur shock ka kya hoga?

Shock ek normal shock ban jaata hai, tangential component

\to 0

, aur deflection

θ \to 0

— strong compression lekin koi turning nahi (Normal Shock Waves).

Jab

β \to μ

(Mach angle) ho toh kya hoga?

M_{n 1} \to 1

, shock ek Mach wave mein weaken ho jaata hai jo koi finite compression carry nahi karta, aur

θ \to 0

(Mach Angle and Mach Waves).

Agar tum Mach angle se chhota

β

plug karo toh?

Formula

M_{1}^{2} sin^{2} β - 1 < 0

deta hai, isliye

tan θ

negative ho jaata hai — physically meaningless; yeh signal karta hai ki

μ

se neeche koi shock exist nahi karta.

Exactly

θ = θ_{m a x}

par kitne solutions hain?

Exactly ek — weak aur strong branches

θ

–

β

curve ke peak par merge ho jaate hain, yeh unique attached shock detachment threshold par hota hai.

Fixed

θ

par

M_{1} \to \infty

pe weak branch ka limiting behaviour kya hai?

Weak wave angle

β

ek finite hypersonic value approach karta hai (zero nahi), isliye shock wedge surface ke karib hota hai chahe Mach angle

μ \to 0

ho.

M_{1} \to \infty

par strong branch ka limiting behaviour kya hai?

Strong-shock

β

9 0^{\circ}

ke karib rehta hai (normal shock approach karta hai) aur

θ_{m a x}

ek finite ceiling approach karta hai (approximately

45. 6^{\circ}

γ = 1.4

ke liye) — ek bahut fast stream zyaada turn ho sakti hai, lekin unboundedly nahi.

Exactly sonic upstream,

M_{1} = 1

par kya hota hai?

Mach angle

μ = sin^{- 1} (1) = 9 0^{\circ}

upper limit par collapse ho jaata hai, isliye sirf allowed

β

9 0^{\circ}

hai aur

θ_{m a x} \to 0

— koi finite turning possible nahi; wedge ek oblique shock se pierce nahi ho sakta, yeh exact threshold hai jiske neeche shocks exist karna band kar dete hain.

Subsonic flow (

M_{1} < 1

) mein ek wedge ka kya hoga?

Koi shock hi nahi — Mach angle defined nahi (

1/ M_{1} > 1

ka koi arcsine nahi hai), signals upstream travel karte hain, aur flow shock ke bajay smoothly adjust ho jaati hai (Supersonic Wedge & Cone Flow).

Agar corner flow ko wedge mein andar ki jagah apne aap se door turn kare (expansion corner), toh kya banta hai?

Shock nahi balki ek smooth Prandtl–Meyer expansion fan — flow ek continuous fan of Mach waves ke through accelerate aur cool hoti hai (Prandtl-Meyer Expansion).

Recall One-line self-test

Sab cover karo: do $β$ -solutions ke naam bolo, unka supersonic/subsonic status, aur $θ_{m a x}$ ke baad unhe kya replace karta hai. Answer ::: Weak (chhota $β$ , downstream supersonic) aur strong (bada $β$ , downstream subsonic); $θ_{m a x}$ ke baad ek detached bow shock form hota hai, koi attached solution nahi hota.