Detached bow shock: Question bank — Hinglish deep dive

3.1.15 · D5 · Physics › Compressible Flow & Aerodynamics › Detached bow shock

Jo vocabulary chahiye woh parent note mein pehle se bani hai: demanded turn == $θ$ , ceiling $θ_{ma x} (M_{1})$ from Maximum deflection angle and weak/strong shock solutions, shock-wave angle $β$ == from Oblique shock waves, Mach angle $μ = arcsin (1/ M_{1})$ , aur centreline normal shock. Kuch naya introduce nahi ho raha — yeh page sirf unhi ideas ko test karta hai. Pehle, un exact symbols ka ek quick refresh jo ye traps use karte hain.

Definition Is page par use hone wale symbols (refresh)

== $Δ$ == — stand-off distance: woh chota sa gap, centreline ke saath measure kiya hua, detached shock ke aage aur body ki naak ke beech. Socho ek curved shock ek blunt nose ke thoda aage latki hui hai; $Δ$ us khaali sliver ki chaudai hai.
== $R_{n}$ == — nose radius: body ki rounded tip ka radius (jaise ek spherical nose ka radius). Stand-off hamesha ratio $Δ/ R_{n}$ ke roop mein batate hain taaki number dimensionless ho aur body-size-independent rahe.
== $ρ_{1}$ , $ρ_{2}$ == — gas ki density shock ke bilkul upstream (subscript 1, shock se pehle) aur bilkul downstream (subscript 2, shock ke baad). Unka ratio $ρ_{2} / ρ_{1}$ normal-shock compression ratio hai; yeh isliye aata hai kyunki mass conservation se $tan (β - θ) / tan β = u_{n 2} / u_{n 1} = ρ_{1} / ρ_{2}$ milta hai.
== $M_{2}$ == — post-shock Mach number: shock ke peeche flow speed ko wahin ki local speed of sound se divide kiya hua. $M_{2} < 1$ matlab shock ke peeche subsonic, $M_{2} > 1$ supersonic, $M_{2} = 1$ exactly sonic.
== $γ$ == — gas ka specific-heat ratio ( $c_{p} / c_{v}$ ). Is page par har numeric value ( $θ_{ma x}$ numbers, compression limit) air with $γ = 1.4$ assume karti hai; alag gas hogi toh sab shift ho jaayenge.

True or false — justify

The shock detaches purely because the body is blunt.

False — bluntness sirf us turn ke zariye matter karta hai jo woh demand karta hai; detachment ki condition hai

θ > θ_{ma x} (M_{1})

, isliye ek blunt body kaafi high

M_{1}

par apne edges ke paas locally attachment support kar sakti hai, aur ek slender body low

M_{1}

par detach kar sakti hai.

A sharp, pointy nose always produces an attached shock.

False — sharpness ek body ko attached shock potentially hold karne deta hai, lekin tabhi jab uska half-angle

\leq θ_{ma x} (M_{1})

rahe; low supersonic Mach par (

M_{1} = 1.5

, aur air mein

γ = 1.4

ke saath

θ_{ma x} \approx 12°

) ek modest cone phir bhi detach kar jaata hai.

Behind a detached bow shock the flow is subsonic everywhere.

False — sirf centreline ke paas wali pocket (sonic line ke andar, jahan

M_{2} < 1

hai) subsonic hai; bahar ki taraf curved shock kamzor padti jaati hai aur uske peeche ka flow (

M_{2} > 1

) phir se supersonic ho jaata hai.

On the stagnation streamline the bow shock behaves as a normal shock.

True — wahaan shock incoming flow ke perpendicular hota hai (

β = 90°

), isliye

M_{n 1} = M_{1} sin 90° = M_{1}

aur har normal-shock relation directly apply hoti hai.

Raising the upstream Mach number always makes the stand-off distance

Δ

larger.

False — ulta hota hai; higher

M_{1}

se stronger compression milti hai, ek denser thinner shock layer banti hai, isliye

Δ/ R_{n}

(nose radius se scale kiya hua stand-off) ek chote finite limit (

\approx 0.143

sphere ke liye air mein) ki taraf sirakta hai aur shock nose ke paas chipak jaati hai.

The θ–β–M relation has no real solution for

β

when the body demands

θ > θ_{ma x}

True — yahi "no real

β

" detachment ka mathematical fingerprint hai: koi bhi attached oblique shock required turn deliver nahi kar sakta, toh physically woh body se pop off ho jaata hai.

A bow shock is a single Mach wave bent around the nose.

False — Mach wave infinitely weak limit hai (

β \to μ

); bow shock ek family of finite-strength obliques hai, centreline par sabse strong (normal) aur sirf bahut door sides par Mach wave mein degenerate hoti hai.

Tangential velocity changes across the bow shock because the shock is curved.

False — har point par shock apne local normal ke saath hi act karta hai, isliye local tangential component wahaan conserve hoti hai; curvature ka bas yahi matlab hai ki normal direction point to point rotate karti hai.

θ = θ_{ma x}

exactly, the shock is comfortably attached.

False — yeh detachment ka razor's edge hai; yeh last attached state hai, aur

θ

mein thodi si bhi upar ki nudge (ya

M_{1}

mein drop) isse bow shock mein tip kar deti hai.

The value

θ_{ma x} \approx 22.97°

M_{1} = 2

holds for any gas.

False —

θ_{ma x}

specific-heat ratio

γ

par depend karta hai; woh number specifically air (

γ = 1.4

) ke liye hai, aur alag

γ

waali gas (jaise monatomic gas with

γ = 5/3

) ka alag ceiling hoga.

Spot the error

"Since

M_{n 1} = M_{1} sin β

and

β = 90°

on the centreline,

M_{n 1}

is at its smallest there."

Error "smallest" mein hai:

sin β

β = 90°

par maximise hota hai, isliye

M_{n 1} = M_{1}

centreline par sabse bada hota hai — yahi exact wajah hai ki centreline shock is family mein sabse strong hai.

"The flow goes subsonic behind the bow shock, so it can never speed back up to supersonic anywhere."

Error curvature ko ignore karta hai: bahar ki taraf jaate hue jaise

β

drop hota hai shock kamzor padti hai, aur sonic line (

M_{2} = 1

) ke baad post-shock flow (

M_{2} > 1

) phir supersonic ho jaati hai — subsonic region local hai, global nahi.

"Because higher Mach means a more violent shock, the shock must sit further out to cope."

Violence aur distance ka yahaan koi sambandh nahi: ek stronger shock ek denser, thinner shock layer produce karta hai, jo shock ko body ke paas kheenchta hai —

Δ

M_{1}

ke saath decrease karta hai.

"Detachment depends only on the body's turning angle

θ

Yeh dono

θ

aur

M_{1}

par depend karta hai (aur numbers ke liye, gas

γ

par bhi), kyunki ceiling

θ_{ma x}

khud

M_{1}

ka function hai; wahi wedge

M_{1} = 2

par detached aur

M_{1} = 4

par attached ho sakti hai.

"Since

θ

rises with

β

, the biggest

β = 90°

gives the biggest turn."

θ

–

β

curve monotone nahi hai:

θ

pehle badhti hai, kisi intermediate

β

par

θ_{ma x}

tak peak karti hai, phir zero par wapas aa jaati hai

β = 90°

par — normal shock flow ko

0°

turn karta hai, sabse zyada nahi.

"An attached oblique and the strong-shock branch are the same solution."

Kisi bhi attachable

θ < θ_{ma x}

ke liye equation do

β

roots deti hai — ek weak aur ek strong solution; usual attached shock weak wali hoti hai, aur woh sirf

θ_{ma x}

par merge karti hain.

Why questions

Why does the flow need to go subsonic behind the nose of a blunt body?

Ek subsonic pocket aage body ko "feel" kar sakti hai aur usके around gently divert ho sakti hai; supersonic flow upstream mein information nahi bhej sakta, isliye us subsonic region ke bina flow blunt nose ko smoothly wrap nahi kar paata.

Why is the shock strongest exactly on the centreline?

Wahaan

β = 90°

hai, isliye poori upstream velocity shock ke normal hai (

M_{n 1} = M_{1}

); shock ka normal Mach number — jo uski strength set karta hai — maximal hai, ise ek full normal shock banata hai.

Why does "no real

β

" translate into a physical shock jumping off the body?

Nature ko phir bhi incoming supersonic flow process karni hai, lekin koi attached oblique geometry required turn satisfy nahi kar sakta; physically ek hi consistent solution bacha hai — ek curved shock jo stand off karta hai, jo centreline turning effectively

90°

handle karne ke liye ek

β = 90°

point reintroduce karta hai.

Why does the tangential velocity being conserved matter for deriving

θ_{ma x}

Yeh oblique shock ko sirf normal component par act karne waale normal shock tak reduce karta hai, giving the clean

tan (β - θ) / tan β = ρ_{1} / ρ_{2}

relation (jahan

ρ_{1}, ρ_{2}

up/downstream densities hain) — wahi equation jiski

θ

mein peak

θ_{ma x}

define karti hai.

Why does the stand-off distance tend to a finite limit rather than zero as

M_{1} \to \infty

Shock layer ki thickness density ratio

ρ_{2} / ρ_{1}

se set hoti hai, jo khud ek strong normal shock par

(γ + 1) / (γ - 1)

par saturate karti hai (air ke liye,

= 6

); finite compression ka matlab finite (chota) gap hai, zero nahi.

Why does raising

M_{1}

increase the turning "capacity"

θ_{ma x}

Faster upstream flow ek bade deflection ke saath stronger shock sustain kar sakta hai solutions khatam hone se pehle, isliye

θ

–

β

curve ka peak chadh jaata hai — air (

γ = 1.4

) ke liye,

M_{1} = 1.5

par

\approx 12°

M_{1} \to \infty

par

\approx 45.6°

tak.

Why do all the

θ_{ma x}

numbers on this page depend on the gas?

Kyunki

γ

, θ–β–M relation mein

M_{1}^{2} (γ + cos 2 β) + 2

denominator ke through enter karta hai;

γ

badlo aur poora

θ

–

β

curve (aur uska peak) shift ho jaata hai, isliye ceiling gas ki bhi property hai, sirf flow ki nahi.

Edge cases

Exactly on the sonic line behind a bow shock, is the flow subsonic or supersonic?

Strictly naa to ek naa doosra — sonic line woh boundary hai jahan

M_{2} = 1

exactly hai; uske andar flow subsonic hai (

M_{2} < 1

), bahar supersonic (

M_{2} > 1

), aur line khud dividing streamsurface hai.

What happens to

β

far out on the wings of the bow shock, at large distance from the nose?

Yeh Mach angle

μ = arcsin (1/ M_{1})

approach karta hai, jahan shock strength vanish ho jaati hai aur yeh ek weak Mach wave mein degenerate ho jaata hai jo flow ko negligibly turn karta hai.

For a wedge sitting right at

θ = θ_{ma x} (M_{1})

, how many oblique-shock solutions exist?

Exactly ek — weak aur strong branches ek single

β

mein merge ho gayi hain;

θ

mein koi bhi increase is last root ko bhi hata deti hai aur detachment force karti hai.

M_{1} \to 1^{+}

(barely supersonic), what happens to

θ_{ma x}

Yeh

0°

ki taraf sirakta hai, kyunki

μ \to 90°

aur poora

θ

–

β

curve collapse ho jaata hai — tab almost koi bhi body itna turning demand karti hai jo possible nahi, isliye Mach 1 ke paas bow shocks hi rule hote hain.

At the extreme edge of a hypersonic bow shock, does the "subsonic pocket / thin dense layer" picture still describe the whole shock?

Nahi — sirf near-centreline region hi hypersonic flow ka thin dense subsonic shock layer hai; far wings weak oblique shocks rehti hain unke peeche supersonic flow ke saath, isliye description nose ke local hai.

If a body's demanded turn is

θ = 0°

(a flat plate aligned with the flow), what shock forms?

Koi bhi finite strength ka nahi — bina kisi turning ke "shock" sirf ek Mach wave hai angle

μ

par; detach karne ke liye kuch hai hi nahi kyunki deflect karne ke liye kuch hai hi nahi.

If the gas were changed from air (

γ = 1.4

) to a monatomic gas (

γ = 5/3

), would the detachment picture change?

Qualitative picture (

θ

mein peak, centreline par normal shock, weakening wings) unchanged hai, lekin har number —

θ_{ma x}

, compression limit

(γ + 1) / (γ - 1)

, stand-off — shift ho jaata hai, isliye quantitative answers naye

γ

ke liye recompute karne padenge.

Recall One-line synthesis

Detachment kabhi bhi akele "blunt vs sharp" ke baare mein nahi hai ::: yeh hamesha single inequality $θ > θ_{ma x} (M_{1})$ hai (jiske numbers gas $γ$ se fix hote hain), jise geometry, Mach number aur gas ek saath padho.