Detached bow shock: Exercises — Hinglish deep dive

3.1.15 · D4 · Physics › Compressible Flow & Aerodynamics › Detached bow shock

Throughout: $γ = 1.4$ for air, aur har symbol ka wahi matlab hai jo parent note mein tha —

$M_{1}$ = upstream (aane wala) Mach number = flow speed ÷ speed of sound,
$θ$ = woh turning angle jo body flow se maangti hai,
$θ_{ma x} (M_{1})$ = sabse bada turn jo us $M_{1}$ par ek attached oblique shock de sakta hai,
$β$ = shock-wave angle (shock aur incoming flow ke beech ka angle),
$M_{n 1} = M_{1} sin β$ = Mach number ka woh hissa jo seedha shock mein jaata hai.

Level 1 — Recognition

L1.1

Woh ek-line rule batao jo decide karta hai ki shock kisi sharp body se attached rehti hai ya bow shock mein detach ho jaati hai.

Recall Solution

Demanded turn ko ceiling se compare karo: $θ > θ_{ma x} (M_{1}) \Rightarrow detached (bow shock), θ \leq θ_{ma x} (M_{1}) \Rightarrow attached.$ $θ_{ma x}$ us $M_{1}$ par $θ$ -vs- $β$ curve ki sabse oonchi peak hai. Uss peak se zyada turning maango toh θ–β–M equation mein koi real $β$ nahi hoga — shock glued nahi reh sakti. Picture ke liye L1.2 dekho.

L1.2

Centreline par (woh stagnation streamline jo seedha nose mein jaati hai), $β$ kya hoga, aur us point ke theek peeche flow subsonic hai ya supersonic?

Recall Solution

Centreline par shock incoming flow ke perpendicular khadi hoti hai, isliye $β = 9 0^{\circ}$ aur $M_{n 1} = M_{1} sin 9 0^{\circ} = M_{1}$ . $β = 9 0^{\circ}$ wali shock normal shock normal shock hoti hai, sabse strong type — yeh hamesha flow ko subsonic kar deti hai ( $M_{2} < 1$ ). Yahi subsonic pocket allow karti hai ki hawa blunt nose ke around gently ghoom sake.

Level 2 — Application

L2.1

$M_{1} = 2.0$ par air ek wedge se milti hai jiska half-angle $θ = 3 0^{\circ}$ hai. Given $θ_{ma x} (2.0) \approx 22.9 7^{\circ}$ , kya shock attach hogi ya detach?

Recall Solution

Compare karo: $3 0^{\circ} > 22.9 7^{\circ}$ . Demand ceiling se zyada hai, toh shock detach ho jaati hai — bow shock banta hai. Mechanically, θ–β–M relation $tan θ = 2 cot β \frac{M _{1}^{2} s i n ^{2} β - 1}{M _{1}^{2} ( γ + c o s 2 β ) + 2}$ ka koi real $β$ nahi hai jo $M_{1} = 2$ par $θ = 3 0^{\circ}$ produce kare.

L2.2

Wohi wedge rakho ( $θ = 3 0^{\circ}$ ) lekin speed badhao $M_{1} = 4.0$ tak, jahan $θ_{ma x} (4.0) \approx 38. 8^{\circ}$ hai. Attach hoga ya detach?

Recall Solution

Ab $3 0^{\circ} < 38. 8^{\circ}$ : demand ceiling se kam hai, toh shock attached aur oblique rehti hai. Faster flow $θ_{ma x}$ badhata hai, flow ko zyada "turning capacity" deta hai. Moral: detachment dono $θ$ aur $M_{1}$ par depend karta hai — speed badhana ek detached shock ko re-attach kar sakta hai.

L2.3

$M_{1} = 3$ par bow shock ki centreline. Normal-shock relation use karke $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{n 1}^{2}}{γ M _{n 1}^{2} - \frac{γ - 1}{2}},$ shock ke peeche $M_{2}$ nikalo.

Recall Solution

Centreline par $β = 9 0^{\circ}$ , toh $M_{n 1} = M_{1} = 3$ . Phir $M_{2}^{2} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{1 + 1.8}{12.6 - 0.2} = \frac{2.8}{12.4} = 0.2258,$ toh $M_{2} = 0.2258 \approx 0.475$ . Subsonic, bilkul waisa hi jaisa normal shock demand karta hai.

Level 3 — Analysis

L3.1

Sphere stand-off correlation $\frac{Δ}{R _{n}} \approx 0.143 exp (\frac{3.24}{M _{1}^{2}})$ use karke $M_{1} = 2$ aur $M_{1} = 4$ par $Δ/ R_{n}$ compute karo, aur trend ko physically explain karo.

Recall Solution

$M_{1} = 2$ par: exponent $= 3.24/4 = 0.81$ , toh $Δ/ R_{n} = 0.143 e^{0.81} = 0.143 (2.248) \approx 0.322$ . $M_{1} = 4$ par: exponent $= 3.24/16 = 0.2025$ , toh $Δ/ R_{n} = 0.143 e^{0.2025} = 0.143 (1.224) \approx 0.175$ . Gap chhota hota jaata hai jaise $M_{1}$ badhta hai. Kyun: zyada $M_{1}$ → stronger compression → denser, thinner shock layer → shock nose ke kareeб baith jaata hai. (Parent mein quoted $\approx 0.41$ at $M_{1} = 2$ thodi alag rounding use karta hai; trend — zyada Mach par chhota $Δ$ — wahi main point hai.)

L3.2

Jab $M_{1} \to \infty$ , $Δ/ R_{n}$ kaunsi limiting value approach karta hai, aur yeh hypersonic shock layers ke baare mein kya batata hai?

Recall Solution

Jab $M_{1} \to \infty$ , exponent $3.24/ M_{1}^{2} \to 0$ , toh $exp (\cdot) \to 1$ aur $\frac{Δ}{R _{n}} \to 0.143.$ Stand-off kabhi zero nahi pohonchti — ek chhote finite floor par settle hoti hai. Physically, hypersonic speeds par shock nose ke paas ek thin, dense shock layer mein chipak jaati hai, lekin nonzero gap hamesha rehta hai kyunki stagnation point par subsonic pocket ka hona zaroori hai taaki flow divert ho sake.

Level 4 — Synthesis

L4.1

$M_{1} = 2.0$ par ek body $θ = 3 0^{\circ}$ demand karti hai (detaches). Tumhe attached chahiye. L2 se pata hai $θ_{ma x} (4.0) \approx 38. 8^{\circ}$ . $M_{1} = 2$ aur $M_{1} = 4$ ke beech, $θ_{ma x} (M_{1})$ monotonically badhta hai. Minimum Mach number estimate karo jis par $θ_{ma x}$ pehli baar $3 0^{\circ}$ tak pohonchta hai, given data points $θ_{ma x} (2.5) \approx 29.8 0^{\circ}$ aur $θ_{ma x} (2.6) \approx 31.0 3^{\circ}$ .

Recall Solution

Hum woh $M_{1}$ chahte hain jahan $θ_{ma x} = 3 0^{\circ}$ ho. Yeh dono given points ke beech hai. $θ_{ma x}$ vs $M_{1}$ par linear interpolation: $M_{1}^{⋆} \approx 2.5 + (2.6 - 2.5) \cdot \frac{30 - 29.80}{31.03 - 29.80} = 2.5 + 0.1 \cdot \frac{0.20}{1.23} \approx 2.5 + 0.0163 \approx 2.516.$ Toh roughly $M_{1} \approx 2.52$ par shock re-attach hoti hai. Isse neeche yeh detached rehti hai; upar, attached. Yeh detachment criterion ko uss fact ke saath jodhta hai ki $θ_{ma x}$ $M_{1}$ ke saath badhta hai.

L4.2

$M_{1} = 3$ lo aur curved bow shock ko centreline se bahar ki taraf follow karo. Centreline par $β = 9 0^{\circ}$ hai (normal shock, $M_{2} \approx 0.475$ ). Bahut door, $β$ Mach angle $μ = arcsin (1/ M_{1})$ approach karta hai. $μ$ compute karo, aur describe karo ki $β = 9 0^{\circ}$ aur $β = μ$ ke beech $M_{2}$ ka kya hota hai.

Recall Solution

Mach angle: $μ = arcsin (1/3) = arcsin (0.3333) \approx 19.4 7^{\circ}$ . $β = 9 0^{\circ}$ se $β = μ \approx 19.4 7^{\circ}$ tak sweep karte hue, $M_{n 1} = M_{1} sin β$ $3$ se $M_{1} sin μ = 3 \cdot (1/3) = 1$ ki taraf girta hai. Shock continuously weak hoti jaati hai:

centreline ke paas $M_{2} < 1$ (subsonic),
sonic line par $M_{2} = 1$ ,
uske baad $M_{2} > 1$ (phir se supersonic),
limit mein $β \to μ$ , $M_{n 1} \to 1$ aur "shock" zero-strength Mach wave mein fade ho jaati hai. Ek curved shock is tarah oblique strengths ki poori family contain karti hai, strongest (normal) se lekar vanishing tak.

Level 5 — Mastery

L5.1

Inverse problem. Bow shock ki centreline normal shock ke peeche tumhara measure kiya hua $M_{2} = 0.50$ hai. Upstream $M_{1}$ recover karo (yani $M_{n 1}$ , kyunki $β = 9 0^{\circ}$ ). Use karo $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{n 1}^{2}}{γ M _{n 1}^{2} - \frac{γ - 1}{2}} .$

Recall Solution

$M_{2}^{2} = 0.25$ set karo aur $x = M_{n 1}^{2}$ ke liye solve karo. $γ = 1.4$ ke saath: $0.25 = \frac{1 + 0.2 x}{1.4 x - 0.2} .$ Cross-multiply karo: $0.25 (1.4 x - 0.2) = 1 + 0.2 x \Rightarrow 0.35 x - 0.05 = 1 + 0.2 x$ . $0.15 x = 1.05 \Rightarrow x = 7.0 \Rightarrow M_{n 1} = 7 \approx 2.646.$ Kyunki $β = 9 0^{\circ}$ , $M_{1} = M_{n 1} \approx 2.65$ . Toh centreline $M_{2} = 0.50$ ka matlab hai incoming Mach approximately $2.65$ hai.

L5.2

Stagnation check. L5.1 ki flow ke liye ( $M_{1} \approx 2.646$ , $M_{2} \approx 0.50$ ), centreline normal shock ke across stagnation-pressure ratio $p_{02} / p_{01}$ nikalo. Use karo $\frac{p _{02}}{p _{01}} = [\frac{( γ + 1 ) M _{n 1}^{2}}{( γ - 1 ) M _{n 1}^{2} + 2}]^{\frac{γ}{γ - 1}} [\frac{γ + 1}{2 γ M _{n 1}^{2} - ( γ - 1 )}]^{\frac{1}{γ - 1}} .$

Recall Solution

$M_{n 1}^{2} = 7$ , $γ = 1.4$ use karo, toh $γ / (γ - 1) = 3.5$ aur $1/ (γ - 1) = 2.5$ . Pehla bracket: $\frac{2.4 \cdot 7}{0.4 \cdot 7 + 2} = \frac{16.8}{4.8} = 3.5$ . Doosra bracket: $\frac{2.4}{2.8 \cdot 7 - 0.4} = \frac{2.4}{19.6 - 0.4} = \frac{2.4}{19.2} = 0.125$ . $\frac{p _{02}}{p _{01}} = 3. 5^{3.5} \cdot 0.12 5^{2.5} .$ $3. 5^{3.5} \approx 80.21$ aur $0.12 5^{2.5} \approx 5.524 \times 1 0^{- 3}$ , toh $\frac{p _{02}}{p _{01}} \approx 80.21 \times 0.005524 \approx 0.443.$ Stagnation pressure ka 44% bachta hai — baaki strong shock mein kho jaata hai. Bow shocks expensive hoti hain: centreline shock jitni strong, total-pressure penalty utni badi.

Flashcards

Attachment condition (one line)?

Shock tab attach hoti hai jab

θ \leq θ_{ma x} (M_{1})

; warna bow shock mein detach ho jaati hai.

Centreline shock angle aur flow regime?

β = 9 0^{\circ}

(normal shock), peeche ki flow subsonic hoti hai.

M_{1} \to \infty

par stand-off floor (sphere correlation)?

Δ/ R_{n} \to 0.143

Shock ke across kya conserved hota hai —

T_{0}

p_{0}

T_{0}

(stagnation temperature);

p_{0}

girti hai kyunki shock irreversible hoti hai.

M_{1} = 3

par Mach angle?

μ = arcsin (1/3) \approx 19.4 7^{\circ}