Question bank — Aerodynamic heating during reentry — stagnation point heat flux Chapman equation

False. ${\dot{q}}_s\propto R_n^{-1/2}$ hai, isliye chhota radius (sharper nose) flux ko badhata hai. Blunt bodies bow shock ko door push karti hain aur boundary layer ko thicken karti hain, surface ko thanda karti hain.

False. Kyunki ${\dot{q}}_s\propto V^3$ hai, $V$ double karne par flux $2^3=8$ se multiply hota hai, 2 se nahi.

False. Density $\sqrt{{\rho }_{\infty }}$ ke roop mein enter karti hai (do competing boundary-layer effects sirf ek square root chhodte hain), isliye ise double karne par flux $\sqrt{2}\approx 1.41$ se multiply hota hai.

False. Peak heating upar hoti hai, ${\rho }_{\text{peak}}=\beta \sin \gamma /3H$ ke paas (yahan $\beta$ ballistic coefficient hai, $\gamma$ flight-path angle, $H$ scale height); $V^3$ factor ek baar deceleration shuru hone par collapse ho jaata hai, isliye product ${\rho }^{1/2}{V}^3$ intermediate altitude par peak karta hai.

False. Yahan flow poori tarah rest par aa jaati hai ($v\to 0$); woh arrested kinetic energy bilkul isliye hai kyunki local temperature aur heat flux maximum hote hain.

True. ${\dot{q}}_s\propto R_n^{-1/2}$ hai, aur $\sqrt{4}=2$ hai, isliye 4× radius aadha flux deta hai.

False. Yeh ek instant/point par instantaneous heat flux (W/m²) deta hai. Total load ke liye poori trajectory par ${\dot{q}}_s$ integrate karna padta hai.

False. Radiative flux ${\dot{q}}_{rad}\propto R_n{\rho }^a{V}^b$ scale karta hai — yeh $R_n$ ke saath badhta hai, convective ke ulta. Super-blunt, convective relief ke badle radiative penalty leta hai.

Error energy content aur transfer rate ko confuse kar raha hai. Enthalpy $V^2$ scale karta hai, lekin wall ke across boundary-layer mass flux ek aur factor $\propto V$ add karta hai (continuity: ${\dot{m}}^{\prime \prime }\approx {\rho }_e{u}_e$), jo transfer rate ko exponent 3 deta hai.

${\rho }_{\infty }$ reentry mein ~5 orders of magnitude span karta hai — jaise sea level par $\approx 1.2{\text{kg/m}}^3$, 60 km par $\approx 4\times 10^{-4}$, 90 km par $\approx 2\times 10^{-5}$. Kyunki flux $\propto \sqrt{{\rho }_{\infty }}$ hai, galat altitude density use karna answer ko constant $C$ thodi galat hone se kahin zyada kharab kar deta hai; hamesha local free-stream density plug karo.

Stagnation point maximum dekhta hai; flux body ke along kum hoti jaati hai jaise flow accelerate hoti hai aur boundary layer grow karti hai. Nose value worst case size karta hai, average nahi.

Constant Earth air ki gas properties ($c_p$, composition, viscosity) encode karta hai. Mars (CO₂) ya doosre atmospheres ko alag $C$ chahiye.

Exponent mein sign error hai. Yeh $R_n^{-1/2}$ hona chahiye; sirf negative exponent hi bada radius flux reduce karta hai, jo bluntness ki actual wajah hai.

Yeh do alag peaks hain. Peak heating high aur early hai (thin air, high speed); peak deceleration deeper aur later hai (dense air, slower speed). Shield aur structure alag-alag worst-case moments face karte hain.

Do factors compound hote hain: available enthalpy $\propto V^2$, times boundary-layer mass flux jo ise deliver karta hai $\propto V$ (continuity se, ${\dot{m}}^{\prime \prime }\approx {\rho }_e{u}_e$). Product $=V^3$.

Bluntness ek detached bow shock surface se door khadi karti hai, jo zyaadatar kinetic energy shocked air mein dump kar deti hai; yeh flow ko bade radius par bhi stretch karti hai, gentler velocity gradient $\beta$ deti hai aur isliye thicker boundary layer jo wall ko insulate karti hai.

$R_n$ stagnation strain rate $\beta \propto V/R_n$ set karta hai (air nose se kitni jaldi accelerate hoti hai), jo boundary-layer thickness $\delta \propto 1/\sqrt{\beta }$ control karta hai. Bada radius → chhota gradient → thicker $\delta$ → kam conductive flux.

Flux $V^3$ par depend karta hai; jis moment braking shuru hoti hai, $V$ (cubed) itni tezi se gir jaata hai ki badhta hua ${\rho }^{1/2}$ compensate nahi kar sakta, isliye flux g-load se pehle/upar peak karta hai, jise dense low air chahiye.

Heat sirf potential gradient ke neeche flow karti hai. Stagnation enthalpy $h_0$ hot gas temperature $T_0$ track karta hai; wall enthalpy $h_w$ thandi $T_w$ track karti hai. Agar wall stagnation gas jitni hot hoti, toh difference (aur flux) zero ho jaata. Boundary layer ke across conduction drive karne wala yahi gap hai.

Extra drag hi point hai — ek blunt body thin air mein upar hi brake karta hai, aur uska low ${\dot{q}}_s$ (bade $R_n$ se) peak heating ko survivable rakhta hai. Streamlining ek tolerable heat problem ko lethal mein trade kar deta.

Radiation shock-heated gas ke glow karne se aata hai: emitted power temperature ke saath steeply badhti hai (aur $T_0\propto V^2$ hai), aur zyada garam gas partially ionize bhi ho jaati hai, isliye zyada speed gas ko linearly se kahin tez heat aur light up dono karti hai — hence bahut bada $b$, aur density se set hone wali glowing gas ki amount se mild $a\sim 1$–$1.5$.

${\dot{q}}_s\to 0$ like $V^3$. Koi relative motion nahi toh koi stagnation compression nahi aur koi convective heating nahi — physics ke saath consistent hai.

Yeh $\sqrt{{\rho }_{\infty }}$ ki tarah 0 ki taraf jaata hai. Koi air nahi matlab heat ya transfer ke liye koi gas nahi, isliye convective flux high speed par bhi vanish ho jaata hai (though radiative aur solar effects is equation ke bahar hain).

Formula ${\dot{q}}_s\to 0$ as $R_n^{-1/2}$ deta hai. Reality mein ek truly flat face flow structure change kar deta hai aur stagnation-point scaling break down ho jaati hai, isliye equation sirf finite, moderately blunt radii ke liye trusted hai.

Full form $\dot{q}\propto (h_0-h_w)$ mein: agar wall enthalpy stagnation enthalpy se zyada ho jaati, toh sign flip ho jaata (wall se heat nikalna). Reentry ke dauran $h_0\gg h_w$ hota hai, isliye yeh strongly positive rehta hai — lekin ek ablating ya radiating hot shield effective gap ko reduce karta hai.

Directly nahi. Yeh ek stagnation-point, blunt-body estimate hai; sharp leading edges ka effective $R_n$ bahut chhota hota hai, local flux enormous hota hai, aur inhe alag (aksar local-flat-plate) heating models chahiye hote hain.

Nahi. Peak heating ${\rho }_{\text{peak}}=\beta \sin \gamma /3H$ par hoti hai (ballistic coefficient $\beta$, flight-path angle $\gamma$, scale height $H$), jo peak deceleration se upar aur pehle hoti hai. Dono extrema time aur altitude dono mein alag hain.