Detached bow shock: Worked examples — Hinglish deep dive

3.1.15 · D3 · Physics › Compressible Flow & Aerodynamics › Detached bow shock

Yeh page Detached bow shock ke liye case-by-case drill room hai. Parent note ne idea bataya tha; yahan hum har tarah ke question ke through grind karte hain jo yeh topic puch sakta hai — "attach vs detach" decision ke har quadrant, degenerate inputs, limiting cases, ek real-world word problem, aur ek exam-style twist.

Kisi bhi number se pehle, ek vaada: har symbol neeche parent note mein earn kiya gaya tha, lekin main har ek ko jab woh aata hai re-anchor karta hoon taaki tumhe kabhi scroll back na karna pade.

Backbone formula (re-anchored)

Neeche sab kuch parent se ek equation par lean karta hai, toh chalte hain ise apne saamne rakhte hain aur ise kabhi mystery ki tarah refer nahi karte.

Scenario matrix

Ek "case" ko inputs ka ek combination samjho jo physics ya method ko change karta hai. Yeh poora grid hai jo yeh topic tumpe throw kar sakta hai.

#	Case class	Kya ise distinct banata hai	Covered by
A	Detaches ( $θ > θ_{ma x}$ )	Demanded turn flow ki capacity se zyada → bow shock	Ex 1
B	Attaches ( $θ < θ_{ma x}$ )	Turn affordable hai → attached oblique shock	Ex 2
C	Exactly at detachment ( $θ = θ_{ma x}$ )	Knife-edge; ek hi shock angle $β$	Ex 3
D	Degenerate body: blunt nose ( $θ \to 90°$ )	Body right-angle turn demand karta hai → hamesha detaches	Ex 4
E	Centreline conditions ( $β = 90°$ )	Bow shock normal shock ki tarah behave karta hai; $M_{2}$ , $p$ , $T$ nikalo	Ex 5
F	Off-centreline point ( $μ < β < 90°$ )	Same curved shock, locally weaker; flow supersonic ho sakta hai	Ex 6
G	Limiting Mach ( $M_{1} \to 1^{+}$ aur $M_{1} \to \infty$ )	Extremes par $θ_{ma x}$ aur stand-off ka behaviour	Ex 7
H	Real-world word problem	Re-entry capsule stand-off distance	Ex 8
I	Exam twist: re-attachment by speeding up	Same body, do Mach numbers, opposite verdicts	Ex 9

Puri jagah use hone wale symbols (sab parent se):

$M_{1}$ = upstream Mach number = flight speed ÷ local speed of sound. "Sound se kitni baar zyada fast."
$M_{2}$ = downstream Mach number = shock ke peeche flow ke liye same ratio.
$θ$ = flow deflection jo body demand karta hai (wedge half-angle, ya blunt nose par $90°$ ).
$θ_{ma x} (M_{1})$ = us $M_{1}$ par attached oblique shock se milne wala sabse bada turn.
$β$ = shock-wave angle, incoming flow ke relative shock ki tilt.
$μ = arcsin (1/ M_{1})$ = Mach angle, possible sabse weak shock tilt (ek Mach wave).
$γ = 1.4$ air ke liye (ratio of specific heats).
$p$ = static pressure = gas jo "push per unit area" lagata hai; subscript 1 = shock se aage, 2 = peeche.
$T$ = static temperature = gas ki random thermal energy ka measure; subscripts 1 (aage) aur 2 (peeche).
$ρ$ = density (Greek letter "rho") = gas ka mass per unit volume; $ρ_{1}$ aage, $ρ_{2}$ peeche. Yeh teen ideal-gas law $p = ρR T$ se tied hain, jahan $R$ gas constant hai.
$Δ$ = shock stand-off distance = detached shock aur nose ke beech ka chhota gap (parent note se).
$R_{n}$ = nose radius = blunt nose ke curvature ka radius (sphere ke liye, uska geometric radius). Hum $Δ$ ko $R_{n}$ ki units mein measure karte hain, yani dimensionless ratio $Δ/ R_{n}$ .

Example 1 — Case A: shock DETACHES

Worked example Statement

Air, $γ = 1.4$ , $M_{1} = 2.0$ . Ek wedge ka half-angle $θ = 30°$ hai. Attach ya detach?

Forecast: calculate karne se pehle guess karo — kya $30°$ Mach 2 ke liye "bada" turn hai? (Hint: Mach 2 sirf mildly supersonic hai.)

Step 1 — Gatekeeper $θ_{ma x} (2.0)$ nikalo. Upar wale θ–β–M relation mein $θ$ ko $β$ par maximize karne se $θ_{ma x} (2.0) \approx 22.97°$ milta hai. Yeh step kyun? $θ_{ma x}$ woh ek number hai jo attach vs detach decide karta hai. Sab kuch isse compare karne par depend karta hai.

Step 2 — Compare karo. $θ = 30° > 22.97° = θ_{ma x}$ . Yeh step kyun? Yeh literally detachment criterion hai $θ > θ_{ma x} \Rightarrow$ detach.

Step 3 — Figure par verdict padho. $M_{1} = 2$ ke liye $θ$ – $β$ curve par, $30°$ par horizontal line peak ke upar baithti hai — koi intersection nahi, toh koi real $β$ exist nahi karta. Shock nose se pop off ho jaata hai: detached bow shock. Yeh step kyun? "No real $β$ " detachment ka mathematical fingerprint hai — equation satisfy nahi ho sakti, toh koi attached shock geometry exist nahi karti.

Verify: $θ = 22.97°$ ko uske peak $β$ par plug karo aur confirm karo ki curve kabhi $30°$ tak nahi pahunchi. Hum numerically VERIFY mein check karte hain ki $M_{1} = 2$ par $max_{β} θ (β) < 30°$ . ✔ Verdict: detach.

Example 2 — Case B: shock ATTACHES

Worked example Statement

Same wedge $θ = 30°$ , lekin ab $M_{1} = 4.0$ .

Forecast: faster flow mein zyada "turning capacity" hoti hai. Kya $30°$ ab bhi zyada hoga?

Step 1 — Naya gatekeeper. $θ_{ma x} (4.0) \approx 38.77°$ . Yeh step kyun? $θ_{ma x}$ $M_{1}$ ke saath badhta hai, toh verdict flip ho sakta hai.

Step 2 — Compare karo. $30° < 38.77°$ → peak ke neeche. Yeh step kyun? Peak ke neeche horizontal line curve ko do baar kaatti hai → do real $β$ exist karte hain (ek weak aur ek strong shock); shock attach ho jaata hai.

Step 3 — Kaun sa $β$ ? Nature almost hamesha weak (smaller- $β$ ) solution pick karta hai. $θ = 30°$ , $M_{1} = 4$ ke liye θ–β–M relation solve karne par weak root $β \approx 45.5°$ milta hai. Yeh step kyun? θ–β–M relation ek $θ$ ke liye do $β$ deta hai; humein batana hoga ki kaun sa physical shock form hota hai, aur special back-pressure conditions ke bahar yeh weak wala hota hai.

Verify: $β = 45.5°$ , $M_{1} = 4$ ko θ–β–M formula mein substitute karo; isse $θ \approx 30°$ wapas milna chahiye. VERIFY mein checked. ✔ Attached, oblique.

Example 3 — Case C: EXACTLY at detachment

Example 4 — Case D: DEGENERATE blunt nose ( $θ \to 90°$ )

Worked example Statement

Ek perfectly blunt (flat-faced ya spherical) nose $M_{1} = 5$ par fly karta hai. Kya yeh kabhi attach hota hai?

Forecast: ek flat face flow ko poora right angle turn karne ke liye kehta hai. Kya koi $θ_{ma x}$ itna bada hota hai?

Step 1 — Body kya turn demand karta hai? Ek blunt nose par stagnation streamline head-on hit karta hai aur shoulder ke around essentially $90°$ divert karna padta hai, toh effectively $θ \to 90°$ . Yeh step kyun? Demanded $θ$ geometry ki property hai, aur blunt face sabse demanding geometry possible hai.

Step 2 — $θ_{ma x}$ ki theoretical ceiling se compare karo. $M_{1} \to \infty$ par bhi, $θ_{ma x}$ air ke liye $\approx 45.6°$ ke paas top out karta hai. Toh $θ_{ma x}$ kabhi $90°$ tak nahi pahunch sakta. Yeh step kyun? Agar $θ_{ma x}$ ki ceiling $\sim 45.6°$ hai, toh koi Mach number blunt body ko attached shock dene nahi deta.

Step 3 — Verdict. Ek blunt nose hamesha ek detached bow shock produce karta hai, kisi bhi supersonic speed par.

Verify: $M_{1} \to \infty$ ( $γ = 1.4$ ) par limiting $θ_{ma x}$ compute karo aur confirm karo ki yeh $90°$ se $≪$ hai. VERIFY mein checked. ✔ Hamesha detaches — yahi reason hai ki hypersonic capsules aur re-entry vehicles mein hamesha bow shocks hote hain.

Example 5 — Case E: CENTRELINE conditions ( $β = 90°$ )

Worked example Statement

$M_{1} = 3.0$ . Bow shock ki stagnation streamline par, $M_{2}$ , pressure ratio $p_{2} / p_{1}$ , aur temperature ratio $T_{2} / T_{1}$ nikalo. (Symbol list se recall karo: $p$ = static pressure, $T$ = static temperature, $ρ$ = density; subscript 1 = shock se aage, 2 = peeche.)

Forecast: centreline par bow shock ek normal shock hai — peeche flow subsonic hona chahiye. Kya $p_{2} / p_{1}$ 3 ke kareeb hoga ya 10 ke?

Step 1 — Normal component. $M_{n 1} = M_{1} sin β = 3 sin 90° = 3$ . Yeh step kyun? Centreline par $β = 90°$ hai, toh shock poori velocity "dekhta" hai — yeh exactly ek normal shock ki tarah behave karta hai.

Step 2 — Downstream Mach. $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{n 1}^{2}}{γ M _{n 1}^{2} - \frac{γ - 1}{2}} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{2.8}{12.4} = 0.2258$ toh $M_{2} = 0.475$ . Yeh step kyun? Yeh standard normal-shock downstream-Mach formula hai; direct substitution.

Step 3 — Pressure ratio. $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{n 1}^{2} - 1) = 1 + \frac{2.8}{2.4} (9 - 1) = 1 + \frac{2.8}{2.4} (8) = 10.33$ Yeh step kyun? Normal shock par pressure jump; $p$ static pressure hai jo gas push karta hai, aur yeh ratio woh number hai jo nose loads aur drag drive karta hai.

Step 4 — Temperature ratio. $\frac{T _{2}}{T _{1}} = \frac{p _{2}}{p _{1}} \cdot \frac{ρ _{1}}{ρ _{2}}, \frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 ) M _{n 1}^{2}}{( γ - 1 ) M _{n 1}^{2} + 2} = \frac{2.4 ( 9 )}{0.4 ( 9 ) + 2} = \frac{21.6}{5.6} = 3.857$ toh $\frac{T _{2}}{T _{1}} = \frac{10.33}{3.857} = 2.679$ . Yeh step kyun? Ideal-gas law $p = ρR T$ pressure $p$ , density $ρ$ aur temperature $T$ ko together tie karta hai; pressure ratio ko density ratio se divide karne par temperature isolate ho jaata hai.

Verify: $M_{2} = 0.475 < 1$ (subsonic ✔). Teeno ratios VERIFY mein numerically checked. Itna bada $T_{2} / T_{1} = 2.68$ exactly yahi reason hai ki nose heating matter karti hai — Stagnation properties across shocks dekho. ✔

Example 6 — Case F: ek OFF-CENTRELINE point

Worked example Statement

Ab bhi $M_{1} = 3.0$ . Bow shock par us point tak bahar move karo jahan local shock angle $β = 40°$ hai. Wahan peeche flow supersonic hai ya subsonic?

Forecast: centreline se door shock weaker hoti hai. Kya hum $M_{2} > 1$ expect karte hain?

Step 1 — Local normal Mach. $M_{n 1} = M_{1} sin β = 3 sin 40° = 3 (0.6428) = 1.928$ . Yeh step kyun? Sirf shock ke normal component ko compress kiya jaata hai; tangential wala unchanged slip ho jaata hai. Yahi poora reason hai ki ek oblique shock "gentler" hota hai.

Step 2 — Downstream normal Mach. $M_{n 2}^{2} = \frac{1 + 0.2 M _{n 1}^{2}}{1.4 M _{n 1}^{2} - 0.2} = \frac{1 + 0.2 ( 3.717 )}{1.4 ( 3.717 ) - 0.2} = \frac{1.743}{5.004} = 0.3484$ toh $M_{n 2} = 0.590$ . Yeh step kyun? Ex 5 jaisa hi normal-shock formula, lekin ab sirf normal Mach par apply hota hai, kyunki shock apne normal ke along act karta hai.

Step 3 — Tangential part restore karo actual $M_{2}$ paane ke liye. Pehle θ–β–M se deflection: $θ (β = 40°, M_{1} = 3) \approx 22.2°$ . Phir $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.590}{s i n ( 40° - 22.2° )} = \frac{0.590}{s i n 17.8°} = \frac{0.590}{0.3057} = 1.93$ Yeh step kyun? $M_{2} > 1$ isliye hai kyunki tangential velocity (shock se untouched) itni badi hai ki normal part sound ke neeche girane ke bawajood total speed supersonic rehti hai.

Verify: $M_{2} = 1.93 > 1$ — supersonic, parent note ke claim ko confirm karta hai ki sonic line ke baad flow re-accelerate karta hai. VERIFY mein checked. ✔

Example 7 — Case G: do LIMITING Mach numbers

Worked example Statement

$θ_{ma x}$ aur stand-off ratio $Δ/ R_{n}$ ko do extremes par describe karo: $M_{1} \to 1^{+}$ aur $M_{1} \to \infty$ .

Forecast: barely-supersonic speed par, flow kitna bada turn afford kar sakta hai? Hypersonic speed par, kya shock door baithti hai ya body se chipak jaati hai?

Step 1 — Low limit $M_{1} \to 1^{+}$ . Mach angle $μ = arcsin (1/ M_{1}) \to arcsin (1) = 90°$ . Poora $θ$ – $β$ curve $β = 90°$ ke paas ek sliver mein squeeze ho jaata hai, toh $θ_{ma x} \to 0$ . Yeh step kyun? Mach 1 se thoda upar flow bilkul bhi turn nahi kar sakta — koi bhi real body detach ho jaata hai. Jaise $θ_{ma x} (1.5) \approx 12°$ , $θ_{ma x} (1.1) \approx 1.5°$ .

Step 2 — High limit $M_{1} \to \infty$ . $θ_{ma x}$ badhta hai lekin $45.6°$ ke paas saturate kar jaata hai (for $γ = 1.4$ ); yeh kabhi $90°$ nahi pahunchta (yeh exactly Case D ki ceiling hai). Yeh step kyun? Yeh ceiling woh reason hai ki blunt bodies (Case D) kisi bhi speed par shock attach nahi kar sakti.

Step 3 — High $M_{1}$ par stand-off. Nose aur shock ke beech ka gap $Δ$ (parent note se) ek sphere ke liye experimentally-fitted correlation follow karta hai, $\frac{Δ}{R _{n}} \approx 0.143 exp (\frac{3.24}{M _{1}^{2}}),$ jahan constants $0.143$ aur $3.24$ first principles se derive nahi hue — yeh spheres ke measured aur computed shock positions ko curve-fit karne se aate hain (classic result jo Billig, 1967 ne report kiya). Jab $M_{1} \to \infty$ toh exponent $3.24/ M_{1}^{2} \to 0$ , toh $\frac{Δ}{R _{n}} \to 0.143 e^{0} = 0.143.$ Yeh step kyun? Humse limiting stand-off maanga gaya hai, aur yeh correlation blunt sphere ke liye standard engineering tool hai; $M_{1} \to \infty$ lene par exponential $1$ par collapse ho jaata hai, sabse chhota finite gap reveal hota hai — hypersonic speed par shock body se chipak jaata hai.

Verify: $θ_{ma x} (1.5) \approx 12°$ , saturation value $\approx 45.6°$ , aur $Δ/ R_{n} \to 0.143$ check karo. VERIFY mein checked. ✔

Example 8 — Case H: REAL-WORLD word problem

Worked example Statement

Ek spherical re-entry capsule ka nose radius $R_{n} = 1.5 m$ hai aur yeh $M_{1} = 6$ par fly karta hai. Shock stand-off distance $Δ$ (metres mein) estimate karo. Phir batao kya hota hai $Δ$ ka agar yeh $M_{1} = 3$ par decelerate kare.

Forecast: Mach 6 (hypersonic) par, kya tum expect karte ho ki shock nose se centimetres ya metres aage hogi?

Step 1 — $M_{1} = 6$ par sphere correlation apply karo. $\frac{Δ}{R _{n}} = 0.143 exp (\frac{3.24}{36}) = 0.143 e^{0.09} = 0.143 (1.0942) = 0.1565.$ Yeh step kyun? Ex 7 ka Billig correlation supersonic flow mein blunt sphere ke liye exactly sahi tool hai, aur $Δ/ R_{n}$ dimensionless hai toh pehle evaluate karna safe hai.

Step 2 — $R_{n}$ se multiply karo. $Δ = 0.1565 \times 1.5 m = 0.2347 m \approx 23.5 cm .$ Yeh step kyun? $Δ/ R_{n}$ ek pure number hai; actual nose radius $R_{n}$ se multiply karne par physical length wapas milti hai (metres × dimensionless = metres).

Step 3 — $M_{1} = 3$ par decelerate karo. $\frac{Δ}{R _{n}} = 0.143 e^{3.24/9} = 0.143 e^{0.36} = 0.143 (1.4333) = 0.2050, Δ = 0.3075 m \approx 30.8 cm .$ Yeh step kyun? Lower Mach → weaker compression → thicker shock layer → bada gap. Jab slow hota hai shock bahar jaata hai.

Verify: Units: $Δ/ R_{n}$ dimensionless hai, metres se multiply karne par metres milte hain. ✔ Dono numbers ( $0.235$ m aur $0.308$ m) VERIFY mein checked, aur $Δ$ badhta hai jab $M_{1}$ girta hai — "higher Mach → smaller stand-off" rule se match karta hai. ✔

Example 9 — Case I: EXAM TWIST — speeding up se re-attach

Worked example Statement

Ek cone ka half-angle $θ = 35°$ hai. Maano ki $M_{1} = 3.0$ par shock detached hai. Woh Mach number roughly nikalo jis par yeh just re-attach karta hai, aur $M_{1} = 5$ par verdict batao.

Forecast: fix yeh hai ki $M_{1}$ badhao jab tak $θ_{ma x} (M_{1})$ $35°$ se aage na badh jaaye. Guess karo: kya woh Mach 3, 4, ya 8 ke aaspas hoga?

Step 1 — Re-attachment condition batao. Re-attachment us pal hoti hai jab $θ_{ma x} (M_{1}) = 35°$ . Us Mach ke neeche, $θ_{ma x} < 35°$ (detached); uske upar, $θ_{ma x} > 35°$ (attached). Yeh step kyun? Re-attachment detachment ka exact reverse hai — humein demanded $θ$ ko badhte hue peak $θ_{ma x} (M_{1})$ ke neeche wapas cross karna hoga.

Step 2 — Crossing bracket karne ke liye $M_{1}$ scan karo. Kuch Mach numbers par θ–β–M curve ka peak evaluate karo:

$θ_{ma x} (3.0) \approx 34.07°$ → $35°$ se thoda neeche → detached (premise se consistent).
$θ_{ma x} (3.2) \approx 35.30°$ → $35°$ se thoda upar → attached.

Toh crossing $M_{1} = 3.0$ aur $3.2$ ke beech hai; interpolate karne par, re-attachment $M_{1} \approx 3.1$ ke paas. Yeh step kyun? Simple root-find: hum $M_{1}$ ko squeeze karte hain jab tak $θ_{ma x}$ $35°$ se pass na ho jaaye.

Step 3 — $M_{1} = 5$ par verdict. $θ_{ma x} (5.0) \approx 41.1° > 35°$ → attached, crossing se comfortably upar. Yeh step kyun? Re-attachment Mach se kaafi upar, flow ki turning capacity easily demanded $35°$ se zyada hai — yeh "Mach badhao to re-attach" wali lesson ek number mein hai.

Answer: shock $M_{1} \approx 3.1$ ke paas re-attach hoti hai, aur $M_{1} = 5$ par yeh firmly attached hai.

Verify: Confirm karo $θ_{ma x} (3.0) \approx 34.1°$ (detached), $θ_{ma x} (3.2) \approx 35.3°$ (attached) — $M_{1} \approx 3.1$ bracket karta hai — aur $θ_{ma x} (5) \approx 41.1°$ . VERIFY mein checked. ✔

Recall Quick self-test (answers cover karo)

Exactly ONE shock angle $β$ kaunse case mein hota hai? ::: Case C, $θ = θ_{ma x}$ (weak aur strong roots peak par merge ho jaate hain). Blunt nose HAMESHA kyun detach karta hai? ::: Yeh $θ \to 90°$ demand karta hai, lekin $θ_{ma x}$ air ke liye $\approx 45.6°$ se kabhi zyada nahi hota, chahe $M_{1} \to \infty$ hi kyun na ho. $M_{1} = 3$ par bow shock ki centreline par flow subsonic hai? ::: Haan, $M_{2} = 0.475 < 1$ (yeh normal shock ki tarah act karta hai, $β = 90°$ ). $M_{1} = 3$ , $β = 40°$ par off-centreline subsonic hai kya? ::: Nahi, $M_{2} \approx 1.93$ — untouched tangential velocity ise supersonic rakhti hai. Capsule $M_{1} = 6$ se $M_{1} = 3$ par slow hone par stand-off $Δ$ kaise change hota hai? ::: Yeh badhta hai ( $\approx 23.5$ cm se $\approx 30.8$ cm tak) — lower Mach, shock bahar jaati hai.