Exercises — Aerodynamic heating during reentry — stagnation point heat flux Chapman equation

KYA karte hain: formula ko symbol by symbol padhte hain.

• $V$ as $V^3$ aata hai → exponent $+3$.
• ${\rho }_{\infty }$ square root ke andar hai, $\sqrt{{\rho }_{\infty }}={\rho }_{\infty }^{1/2}$ → exponent $+\frac{1}{2}$.
• $R_n$ as $1/\sqrt{R_n}=R_n^{-1/2}$ aata hai → exponent $-\frac{1}{2}$.

Answer: ${\dot{q}}_s\propto V^3{\rho }_{\infty }^{1/2}{R}_n^{-1/2}$.

${\dot{q}}_s$ ek heat flux hai: energy per unit time per unit area, isliye ${\text{W/m}}^2={\text{J⋅s}}^{-1}{\text{⋅m}}^{-2}$. Yeh measure karta hai how fast heat pours into each square metre of the nose surface. Yeh total energy nahi hai — yeh ek rate per area hai.

Stagnation point — nose ka woh point jahan incoming flow puri tarah rest mein aa jaati hai ($v\to 0$). Saari ordered kinetic energy local enthalpy (temperature/pressure) mein convert ho jaati hai, isliye is point par maximum convective heat flux hoti hai. Dekhein Bow shock and blunt-body theory yeh samajhne ke liye ki blunt nose yahan detached shock kyun banata hai.

Pieces ko factor kyun karein? Taaki har square root aur cube honest rahe.

• $\sqrt{R_n}=\sqrt{1.0}=1.0$.
• $\sqrt{{\rho }_{\infty }}=\sqrt{5.0\times 10^{-4}}=2.236\times 10^{-2}$.
• $V^3=(7500{)}^3=4.219\times 10^{11}$.

${\dot{q}}_s=\frac{1.83\times 10^{-4}}{1.0}\times 2.236\times 10^{-2}\times 4.219\times 10^{11}$ $=1.83\times 10^{-4}\times 2.236\times 10^{-2}\times 4.219\times 10^{11}\approx 1.726\times 10^6{\text{W/m}}^2.$ Convert: $1{\text{W/cm}}^2=10^4{\text{W/m}}^2$, isliye ${\dot{q}}_s\approx 173{\text{W/cm}}^2$.

KYA govern karta hai: sirf $V^3$ factor move karta hai. KYun matter karta hai: yeh reentry heating mein sabse bada lever hai. $\frac{{\dot{q}}_s(2V)}{{\dot{q}}_s(V)}=\frac{(2V{)}^3}{V^3}=2^3=8.$ Speed double karne se heating 8 guna badtar ho jaati hai.

Sirf $R_n^{-1/2}$ move karta hai. $\frac{{\dot{q}}_s(4R_n)}{{\dot{q}}_s(R_n)}=\frac{(4R_n{)}^{-1/2}}{R_n^{-1/2}}=4^{-1/2}=\frac{1}{2}.$ 4 guna bada nose radius heat flux ko aadha kar deta hai — isliye capsules blunt hote hain. Dekhein Thermal Protection Systems (ablatives, tiles).

Ratio kyun lein? Shared ${\rho }_{\infty }$ aur $C$ cancel ho jaate hain, isliye sirf $V^3$ aur $R_n^{-1/2}$ pieces bachte hain. $\frac{{\dot{q}}_A}{{\dot{q}}_B}={\left(\frac{V_A}{V_B}\right)}^3{\left(\frac{R_{n,B}}{R_{n,A}}\right)}^{1/2}$

• Speed part: $(8000/6000{)}^3=(4/3{)}^3=64/27=2.370$.
• Radius part: $(2.0/0.5{)}^{1/2}=4^{1/2}=2.0$.

$\frac{{\dot{q}}_A}{{\dot{q}}_B}=2.370\times 2.0=\mathrm{4.741.}$ Vehicle A lagbhag $4.74\times$ zyada garam hota hai — faster bhi hai aur bluntness mein disadvantaged bhi.

KYA product jaisa dikhta hai: ${\dot{q}}_s$ ek badhte factor ${\rho }^{1/2}$ aur ek girte factor $V^3$ ka product hai jab vehicle neeche aata hai. Figure dekhein.

• Atmosphere ke top ke paas: $\rho \to 0$, isliye ${\rho }^{1/2}\to 0$ → almost koi heat nahi chahe $V$ kitna bhi bada ho.
• Bottom ke paas: vehicle ne hard brake lagayi hai, $V$ chhota hai, aur $V^3$ (cube hone ki wajah se!) collapse kar jaata hai → heat phir se girta hai.
• Beech mein, product ek intermediate altitude par maximise hota hai. Kyunki $V$ cube mein enter karta hai aur $\rho$ sirf half-power mein, $V^3$ collapse dominate karta hai jaise hi braking shuru hoti hai, isliye peak upar, well before peak deceleration ke baith jaata hai (dekhein Ballistic coefficient and deceleration).

• $\sin 30^{\circ }=0.5$.
• ${\rho }_{\text{peak}}=\frac{200\times 0.5}{3\times 7000}=\frac{100}{21000}=4.76\times 10^{-3}{\text{kg/m}}^2\cdot {\text{m}}^{-1}$.

Ruko — dimensions check karo: $\beta$ yahan ballistic coefficient ${\text{kg/m}}^2$ hai, aur length $H$ (m) se divide karne par ${\text{kg/m}}^3$ milta hai, jo ek density hai. Isliye ${\rho }_{\text{peak}}\approx 4.76\times 10^{-3}{\text{kg/m}}^3$. Sea level ($1.2{\text{kg/m}}^3$) se compare karein, yeh bahut thin, high-altitude density hai — L3.2 confirm karta hai: peak heating upar hoti hai. Dekhein Reentry trajectory dynamics.

Factors multiply kyun karein? Har variable ek independent power mein enter karta hai, isliye unke effects multiply hote hain. $\frac{{\dot{q}}_{\text{new}}}{{\dot{q}}_{\text{old}}}=(0.9{)}^3\times (2{)}^{-1/2}$

• $(0.9{)}^3=0.729$.
• $2^{-1/2}=0.7071$.

$\frac{{\dot{q}}_{\text{new}}}{{\dot{q}}_{\text{old}}}=0.729\times 0.7071=\mathrm{0.5155.}$ Net heat flux lagbhag 51.6% tak gir jaata hai — almost aadha. Speed reduction (cubed) zyada kaam karta hai, bluntness baaki ka kaam karta hai.

KYA karte hain: formula ko $R_n$ ke liye invert karte hain. KYun invert karte hain: unknown $1/\sqrt{R_n}$ ke andar bura hua hai, isliye hum ise algebraically isolate karte hain. ${\dot{q}}_s=\frac{C\sqrt{{\rho }_{\infty }}{V}^3}{\sqrt{R_n}}\le {\dot{q}}_{\max }\Rightarrow \sqrt{R_n}\ge \frac{C\sqrt{{\rho }_{\infty }}{V}^3}{{\dot{q}}_{\max }}.$ Pehle numerator compute karo:

• $\sqrt{{\rho }_{\infty }}=\sqrt{8.0\times 10^{-4}}=2.828\times 10^{-2}$.
• $V^3=(9000{)}^3=7.29\times 10^{11}$.
• Numerator $=1.83\times 10^{-4}\times 2.828\times 10^{-2}\times 7.29\times 10^{11}=3.773\times 10^6$.

$\sqrt{R_n}\ge \frac{3.773\times 10^6}{5.0\times 10^6}=0.7546\Rightarrow R_n\ge 0.7546^2=0.569\text{m}.$ Minimum nose radius $\approx 0.57\text{m}$.

(a) Pieces factor karo:

• $\sqrt{R_n}=\sqrt{4.7}=2.168$.
• $\sqrt{{\rho }_{\infty }}=\sqrt{10^{-4}}=10^{-2}$.
• $V^3=(11000{)}^3=1.331\times 10^{12}$.

${\dot{q}}_s=\frac{1.83\times 10^{-4}}{2.168}\times 10^{-2}\times 1.331\times 10^{12}=8.44\times 10^{-5}\times 10^{-2}\times 1.331\times 10^{12}$ $\approx 1.12\times 10^6{\text{W/m}}^2\approx 112{\text{W/cm}}^2.$ Lunar return ke liye sahi order hai. ✔

(b) Convective: ${\dot{q}}_{\text{conv}}\propto R_n^{-1/2}$ → bada nose ⇒ thicker boundary layer, gentler velocity gradient ⇒ kam convective flux. Radiative: shock-layer radiation roughly ${\dot{q}}_{\text{rad}}\propto R_n{\rho }^a{V}^b$ scale karti hai ($R_n$ ka positive power) — ek bada, thicker glowing shock layer zyada radiate karta hai zyada. Isliye bluntness convective relief ko radiative penalty se trade karti hai; real designs dono ko balance karte hain.

Convective ($\propto V^3$): ${\dot{q}}_{\text{conv}}^{\prime }=110\times (1.1{)}^3=110\times 1.331=146.4{\text{W/cm}}^2$. Radiative ($\propto V^{8.5}$): ${\dot{q}}_{\text{rad}}^{\prime }=40\times (1.1{)}^{8.5}$. Ab $(1.1{)}^{8.5}=e^{8.5\ln 1.1}=e^{8.5\times 0.09531}=e^{0.8101}=2.248$. Isliye ${\dot{q}}_{\text{rad}}^{\prime }=40\times 2.248=89.9{\text{W/cm}}^2$.

Interpretation: Sirf $10\%$ speed bump convective heating ko $\sim 33\%$ badhata hai lekin radiative heating ko double se zyada kar deta hai, kyunki iska exponent ($\sim 8.5$) convective $3$ ko dwarf karta hai. Yahi reason hai ki sabse zyada reentry speeds par radiation dominant threat ban jaata hai — bade $b$ wala $V^b$ $V^3$ ko overtake kar leta hai.

Formula ${\dot{q}}_s=\frac{C}{\sqrt{R_n}}\sqrt{{\rho }_{\infty }}{V}^3$; $V$ cubed (double $V$ ⇒ 8×), ${\rho }_{\infty }$ half-power, $R_n^{-1/2}$ (4× radius ⇒ half flux). Peak heating upar hoti hai (thin air mein), peak deceleration se pehle. Radiation $R_n$ ka sign reverse karta hai aur ek bada $V^b$ use karta hai.