Worked examples — Hypersonic flow — Mach 5+, high temperature effects

3.1.27 · D3 · Physics › Compressible Flow & Aerodynamics › Hypersonic flow — Mach 5+, high temperature effects

Shuru karne se pehle, ek quick tool inventory (har ek parent mein define tha — hum sirf use karte hain):

Recall Woh paanch formulas jinpar hum tike rahenge

• Stagnation temperature: $\frac{T_0}{T}=1+\frac{\gamma -1}{2}{M}^2$
• Mach number: $M=\frac{V}{a}$, with speed of sound $a=\sqrt{\gamma RT}$
• Limiting density ratio (normal shock): $\frac{{\rho }_2}{{\rho }_1}\to \frac{\gamma +1}{\gamma -1}$
• Degrees of freedom: $\gamma =1+\frac{2}{f}$
• Newtonian pressure coefficient (unified): $C_p=2{\sin }^2\theta$ for $\theta >0$ (surface flow ki taraf face kar rahi hai), aur $C_p=0$ for $\theta \le 0$ (surface shadow mein hai). Hum inhe ek formula plus ek patch nahi, balki ek rule ke do branches mante hain.

Poore note mein, air ke liye $R=287{\text{J kg}}^{-1}{\text{K}}^{-1}$ hai aur (jab calorically perfect ho) $\gamma =1.4$.

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Scenario matrix

Is chapter ka har hypersonic problem inhi cells mein se ek hai. Neeche ke examples mein [Cell #] tag carry karte hain taaki tum dekh sako ki poora space cover ho raha hai.

Cell	Case class	Kya tricky banata hai	Example
1	Moderate $M$ (just past 5)	plug-and-chug, lekin temperature sanity-check karo	Ex 1
2	Large finite $M$ (re-entry)	perfect-gas answer ek overestimate hai	Ex 2
3	Limiting input $M\to \infty$	formula ko ek finite, meaningful cap dena chahiye	Ex 3
4	Degenerate angle $\theta =0$	grazing flow, $C_p$ smoothly vanish hona chahiye	Ex 4
5	Negative / shadow angle $\theta <0$	leeward face — $C_p=0$ return karna chahiye, negative nahi	Ex 5
6	Full angle sweep (windward face)	check karo $\theta =90^{\circ }$ stagnation cap deta hai	Ex 6
7	Real-gas twist (small $\gamma$)	same $M$, different $\gamma$ → bahut alag physics	Ex 7
8	Density-ratio limit with real gas	shock layer kyun collapse karta hai	Ex 8
9	Word problem (re-entry heat budget)	sahi tool chuno, units honest rakho	Ex 9
10	Exam twist (Mach-independence)	trap: "kya Mach double karne se force double hoti hai?"	Ex 10

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Ex 1 — Moderate Mach, threshold se thoda aage [Cell 1]

Forecast: compute karne se pehle guess karo — kya $T_0$ 500 K, 1000 K, ya 1500 K ke aas-paas hogi?

1. Ratio likho. $\frac{T_0}{T}=1+\frac{\gamma -1}{2}{M}^2$. Yeh step kyun? Yeh steady-adiabatic energy equation hai jo stagnation temperature ke liye solve ki gayi hai — yahi ek tool hai jo "kitni tez" ko "kitni garm" mein convert karta hai.
2. Numbers daalo. $\frac{\gamma -1}{2}=\frac{0.4}{2}=0.2$, aur $M^2=25$, isliye $\frac{T_0}{T}=1+0.2\times 25=6$. Yeh step kyun? Right side par sab kuch ab ek pure number hai; ratio $6$ hai.
3. Multiply back karo. $T_0=6\times 250=1500\text{K}$. Yeh step kyun? Hume ek temperature chahiye thi, isliye free-stream $T$ se un-divide kiya.

Verify: exactly $M=5$ par ratio $1+0.2(25)=6$ hai, ek clean small integer — achha sign hai. $1500\text{K}$ ${\text{O}}_2$ dissociation ki doorstep par hai, yahi wajah hai ki $M\approx 5$ ko hypersonic threshold kaha jaata hai. Forecast answer teesra option hai.

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Ex 2 — Large Mach: jahan perfect-gas number jhooth bolta hai [Cell 2]

Forecast: kya perfect-gas $T_0$ 10 000 K, 20 000 K, ya 40 000 K ke kareeb hogi?

1. Perfect-gas prediction. $\frac{T_0}{T}=1+0.2\times (20{)}^2=1+0.2\times 400=81$. Yeh step kyun? Ex 1 jaisi hi formula — hum ise deliberately apne comfort zone se bahar apply kar rahe hain error expose karne ke liye.
2. Multiply back karo. $T_0=81\times 250=20250\text{K}$. Yeh step kyun? Yeh Sun ki surface se bhi zyada garm hai — ek red flag ki model broken hai.
3. Physical correction. $\sim 2500\text{K}$ se upar air dissociate hoti hai: energy thermal pool se nikal kar O${}_2$ aur N${}_2$ bonds todne mein jaati hai (dekho Real Gas Thermodynamics & Dissociation). Yeh step kyun? Total enthalpy conserve hota hai, lekin ab woh chemistry mein stored hai, temperature mein nahi — isliye real $T$ sirf $\sim 6000\text{–}8000\text{K}$ ho sakti hai.

Verify: ratio $81$ hai $1+0.2M^2$ at $M=20$; check karo $0.2\times 400=80$, plus $1$ deta hai $81$ — arithmetic theek hai. Lesson: number mathematically correct hai lekin physically ek overestimate hai, exactly wahi [!mistake] jo parent note mein warn kiya gaya tha.

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Ex 3 — Limiting input $M\to \infty$ [Cell 3]

Forecast: kya ${\rho }_2/{\rho }_1$ without bound badhta hai, ya ek finite ceiling par settle hota hai?

1. Limit state karo. Rankine–Hugoniot Relations se, jab $M\to \infty$, $\frac{{\rho }_2}{{\rho }_1}\to \frac{\gamma +1}{\gamma -1}$. Yeh step kyun? Temperature ke unlike (jo $M^2$ ke saath badhti hai), density capped hai — mass conservation aur momentum jump infinite compression forbid karte hain.
2. $\gamma =1.4$ par evaluate karo. $\frac{1.4+1}{1.4-1}=\frac{2.4}{0.4}=6$. Yeh step kyun? Cold air ke liye yeh famous factor-of-6 ceiling hai.
3. $\gamma =1.1$ par evaluate karo (real-gas softened). $\frac{1.1+1}{1.1-1}=\frac{2.1}{0.1}=21$. Yeh step kyun? Real-gas effects $\gamma$ ko shrink karte hain, isliye same infinite Mach gas ko ~3.5× zyada squeeze karta hai → aur bhi 얇얇얇얇thinner shock layer.

Verify: $\gamma =1.4$ plug karo: numerator $2.4$, denominator $0.4$, ratio $6.0$ ✓. Trend sanity check: smaller $\gamma -1$ denominator mein ⇒ bigger ratio ⇒ $21>6$ ✓. Answer finite hai (pehla forecast galat hai): density kabhi diverge nahi karti.

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Ex 4 — Degenerate angle $\theta =0$ (grazing flow) [Cell 4]

Forecast: zero, ya koi small positive residual?

Figure — Ex 4 (grazing flow): lavender arrows stream ko solid body ke parallel run karte dikhate hain; ek coral note mark karta hai ki koi arrow wall ke andar point nahi karta, isliye koi normal momentum deliver nahi hota. Yahi geometry hai jo $\theta =0$ ko unified rule ke do branches ke beech ki boundary banati hai.

1. Geometry padho. Figure mein incoming particles (lavender arrows) surface ke along move karte hain; unki momentum mein se koi bhi wall ke andar point nahi karta. Yeh step kyun? Newtonian pressure normal momentum se aata hai jo destroy hota hai — agar normal component zero hai, toh koi momentum deliver nahi hota.
2. Boundary par unified rule apply karo. $\theta =0$ par dono branches agree karte hain: windward branch deta hai $2{\sin }^2{0}^{\circ }=0$, aur shadow branch deta hai $0$. Isliye $C_p=0$. Yeh step kyun? $\theta =0$ exactly "flow ki taraf face karna" aur "shadow mein hona" ke beech ka seam hai; ek accha rule wahan continuous hona chahiye, aur yeh rule hai.

Verify: units-free coefficient, aur $2{\sin }^2(0)=0$ ✓. Limit smooth hai: tiny angle $\theta =1^{\circ }$ ke liye, $C_p=2{\sin }^2{1}^{\circ }\approx 0.00061$ — almost zero, confirming no jump.

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Ex 5 — Shadow / negative angle $\theta <0$ (leeward face) [Cell 5]

Forecast: kya formula ek negative $C_p$ deta hai, ya hum ise override karte hain?

Figure — Ex 5 (shadow face): wedge ek mint windward face (lavender stream se struck) aur ek coral leeward face present karta hai jo flow se away point karta hai. Coral annotation unified rule ke $\theta \le 0$ branch ko mark karta hai, jahan model zero impacts predict karta hai aur isliye $C_p=0$.

1. Shadow spot karo. Figure mein, coral face stream se away face karta hai: koi bhi particle ise reach nahi kar sakta (얇thin shock layer sirf front wrap karta hai). Yeh step kyun? Newtonian theory ek particle-impact model hai — zero impacts receive karne wali surface zero excess pressure feel karti hai, by construction.
2. Sahi branch select karo — koi override needed nahi. Kyunki $\theta =-20^{\circ }\le 0$ hai, unified rule khud hume $C_p=0$ deta hai. Hum $2{\sin }^2\theta$ yahan evaluate hi nahi karte; woh branch simply apply nahi hoti. Yeh step kyun? Single signed rule mein shadow case already contained hai, isliye kuch bhi ad hoc patch nahi karna — hum sirf $\theta$ ke sign se selected branch padh lete hain.
3. Blind squaring kyun mislead karta. Agar koi branching bhool jaaye aur $2{\sin }^2\theta$ mein plug kare, toh use $2{\sin }^2(-20^{\circ })=2{\sin }^2(20^{\circ })\approx 0.234$ milega, kyunki ${\sin }^2$ even hai aur sign discard kar deta hai. Unified rule isko prevent karta hai pehle sign read karke. Yeh step kyun? Yeh precisely woh mistake dikhata hai jiske against branching guard karta hai.

Verify: physical answer $C_p=0$ hai (sahi branch), not $0.234$. Sanity check: shadowed face par pressure free-stream $p_{\infty }$ par baith hai, yaani $p-p_{\infty }=0\Rightarrow C_p=0$ ✓.

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Ex 6 — Full windward sweep: $\theta =90^{\circ }$ cap [Cell 6]

Forecast: is theory ke according sabse bada possible $C_p$ guess karo.

Figure — Ex 6 ($C_p$ sweep): lavender curve windward branch $C_p=2{\sin }^2\theta$ ko $0$ se $90^{\circ }$ tak plot karta hai. Teen highlighted dots (mint, butter, coral) hamare teen angles mark karte hain, aur dashed coral ceiling stagnation cap $C_p=2$ dikhata hai jo curve $\theta =90^{\circ }$ par reach karta hai.

1. $\theta =15^{\circ }$. $C_p=2{\sin }^{2}15^{\circ }=2(0.2588{)}^2=0.134$. Yeh step kyun? Parent note ke flat-plate example se match karta hai — ek accha cross-check ki hamare tool sahi se wired hain.
2. $\theta =45^{\circ }$. $C_p=2{\sin }^{2}45^{\circ }=2(0.7071{)}^2=2(0.5)=1.0$. Yeh step kyun? Angle mein half-way, lekin ${\sin }^2$ nonlinear hai, isliye hum exactly $1$ par land karte hain.
3. $\theta =90^{\circ }$ (surface flow ke square — stagnation point). $C_p=2{\sin }^{2}90^{\circ }=2(1{)}^2=2$. Yeh step kyun? Yeh maximum windward $C_p$ hai; poora normal momentum destroy hota hai. Figure mein curve yahan peak karta hai.

Verify: $2{\sin }^{2}90^{\circ }=2$ ceiling hai kyunki ${\sin }^2\le 1$ ✓. Monotone rise $0.134\to 1.0\to 2$ figure ke climbing curve se match karta hai ✓. Forecast maximum $C_p=2$ hai.

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Ex 7 — Real-gas twist: same Mach, softer $\gamma$ [Cell 7]

Forecast: kya smaller $\gamma$ predicted ratio ko raise ya lower karta hai?

1. $\gamma =1.4$ case. $\frac{T_0}{T}=1+\frac{0.4}{2}(25)=1+5=6$. Yeh step kyun? Ex 1 se baseline — hum ise reuse karte hain taaki sirf $\gamma$ change ho.
2. $\gamma =1.2$ case. $\frac{\gamma -1}{2}=\frac{0.2}{2}=0.1$, isliye $\frac{T_0}{T}=1+0.1(25)=1+2.5=3.5$. Yeh step kyun? Smaller $\gamma$ ka matlab zyada degrees of freedom jo energy soak up karte hain, isliye same kinetic energy temperature kam raise karti hai — ratio $6$ se $3.5$ par aa jaata hai.
3. $f$ ke through interpret karo. $\gamma =1+2/f$ se: $\gamma =1.4\Rightarrow f=5$; $\gamma =1.2\Rightarrow f=2/(0.2)=10$. Yeh step kyun? Dس active "storage drawers" vs panch — energy hide karne ke liye do baar jagah, isliye bahut lower temperature. Real Gas Thermodynamics & Dissociation se consistent.

Verify: $1+0.1\times 25=3.5$ ✓ aur $f=2/(\gamma -1)=2/0.2=10$ ✓. Trend check: lower $\gamma$ ⇒ lower $T_0/T$ ✓ (forecast "lower" sahi hai).

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Ex 8 — Real-gas $\gamma$ ke saath density-ratio limit [Cell 8]

Forecast: cold air ki limit se roughly kitna zyada?

1. Limit apply karo. $\frac{{\rho }_2}{{\rho }_1}\to \frac{\gamma +1}{\gamma -1}=\frac{2.2}{0.2}=11$. Yeh step kyun? Same $M\to \infty$ normal-shock limit jaise Ex 3 mein, ab hot-gas $\gamma$ ke saath.
2. Compare karo. Cold air $6$ par cap hoti hai; hot gas $11$ tak pahunchti hai, factor $11/6\approx 1.83$. Yeh step kyun? Zyada compression ka matlab same post-shock mass kam volume occupy karta hai.
3. Geometric consequence. Zyada density ⇒ 얇얇얇얇얇thinner shock layer ⇒ bow shock body ke aur kareeb hug karta hai, wall heating problem ko intensify karta hai. Yeh step kyun? Mass conservation density ko layer thickness se directly tie karta hai.

Verify: $\frac{2.2}{0.2}=11$ ✓ aur $11/6=1.833\dots$ ✓. Forecast "roughly 2×" sahi hai.

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Ex 9 — Word problem: re-entry heat budget [Cell 9]

Yahan $T_{\infty }$, $a_{\infty }$, $V_{\infty }$ sab subscript $\infty$ carry karte hain yaani "undisturbed free stream mein," body se door. Sound-speed formula poore mein same free-stream temperature $T_{\infty }$ use karta hai — koi symbol switching nahi.

Forecast: kinetic energy per kilogram 1 MJ, 10 MJ, ya 30 MJ ke aas-paas hogi?

1. Free-stream speed of sound. $a_{\infty }=\sqrt{1.4\times 287\times 250}=\sqrt{100450}\approx 316.9\text{m/s}$. Yeh step kyun? Mach number ek ratio hai; real speed pane ke liye hume local (yahan free-stream) sound speed chahiye — woh yardstick jiske against $M$ measure hota hai.
2. Free-stream speed. $V_{\infty }=M{a}_{\infty }=25\times 316.9\approx 7923\text{m/s}$. Yeh step kyun? $M=V_{\infty }/a_{\infty }$ rearrange kiya; $\sim 7.9\text{km/s}$ ek realistic orbital re-entry speed hai — ek accha reality anchor.
3. Kinetic energy per kg. $\frac{1}{2}{V}_{\infty }^2=0.5\times (7923{)}^2\approx 3.14\times 10^7\text{J/kg}=31.4\text{MJ/kg}$. Yeh step kyun? Steady energy equation ke anusaar yahi exactly woh enthalpy hai jo gas ko stagnation par absorb karni padti hai — "heat budget."
4. Kahan jaati hai. Sirf ek part temperature banta hai; baaki bonds todne mein (dissociation) aur electrons lift karne mein (ionization) jaata hai. Wall heating aur radio blackout dono is pool se aate hain. Yeh step kyun? Number ko parent note mein physics ladder se tie karta hai.

Verify: $a_{\infty }=\sqrt{1.4\cdot 287\cdot 250}\approx 316.9$ m/s ✓; $V_{\infty }=25a_{\infty }\approx 7923$ m/s ✓; $\frac{1}{2}{V}_{\infty }^2\approx 3.14\times 10^7$ J/kg ✓. Units: $(\text{m/s}{)}^2={\text{m}}^2/{\text{s}}^2=\text{J/kg}$ ✓. Forecast "≈30 MJ" sahi hai.

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Ex 10 — Exam twist: kya Mach double karne se force double hoti hai? [Cell 10]

Forecast: factor 1 (unchanged), factor 2, ya factor 4?

1. $M=8$ par $C_p$ compute karo. $\theta =15^{\circ }>0$ ke saath hum windward branch lete hain: $C_p=2{\sin }^{2}15^{\circ }=0.134$. Yeh step kyun? Scaling ke baare mein reasoning karne se pehle hum actual unified rule evaluate karte hain.
2. $M=16$ par $C_p$ compute karo. Phir bhi $C_p=2{\sin }^{2}15^{\circ }=0.134$ — $M$ rule mein appear hi nahi karta. Yeh step kyun? Yeh Mach-independence principle hai: hypersonic speed par pressure coefficient sirf geometry ($\theta$) par depend karta hai, Mach number par nahi.
3. Trap diagnose karo. Student ne $C_p$ (ek coefficient, already $\frac{1}{2}{\rho }_{\infty }{V}_{\infty }^2$ se normalized) ko dimensional pressure ke saath confuse kiya. Pressure $p-p_{\infty }$ does $V_{\infty }^2\propto M^2$ ke saath scale karta hai, isliye woh $\times 4$ badhta hai — lekin coefficient unchanged hai. Yeh step kyun? Coefficient ko raw force se alag karna hi is exam twist ka pura point hai.

Verify: $C_p(M=8)=C_p(M=16)=0.134$, ratio $=1$ ✓. Jabki $p-p_{\infty }\propto V_{\infty }^2$, isliye $M$ (aur hence $V_{\infty }$) double karne se pressure multiply hota hai $2^2=4$ se ✓. Coefficient question ka sahi answer factor 1 hai — student galat tha.

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Recall Matrix ke across quick self-test

$M=5$, $\gamma =1.4$ par, $T_0/T$ kya hai? ::: $6$. $M\to \infty$ jab $\gamma =1.4$, shock density ratio kis par cap hoti hai? ::: $6$. Shadowed (leeward, $\theta \le 0$) face par $C_p$? ::: $0$ (unified rule ka shadow branch). Maximum windward $C_p$ (surface flow ke square, $\theta =90^{\circ }$)? ::: $2$. $M$ double karne se $C_p$ kis factor se change hota hai? ::: $\times 1$ (unchanged — Mach independence). $M$ double karne se actual pressure $p-p_{\infty }$ kitne se change hota hai? ::: $\times 4$ (yeh $V_{\infty }^2$ ke saath scale karta hai).