Thermal protection systems — ablators (PICA, SLA), metallic tiles, RCC: Worked examples

3.4.22 · D3 · Physics › Rocket Flight Mechanics › Thermal protection systems — ablators (PICA, SLA), metallic

Koi bhi number se pehle, teen tools ko simple words mein dobara state karte hain taaki har symbol earn kiya gaya ho.

Formula Teen tools (zero se dobara stated)

Tool 1 — Convective heating (Sutton–Graves). $\overset{q}{˙}_{conv} = k \frac{ρ}{R _{n}} v^{3}$

$\overset{q}{˙}_{conv}$ = heat power jo nose ke ek square metre par aa raha hai, watts per square metre mein ( $W/m^{2}$ ). Socho "ek stamp-sized patch par heat kitni tezi se gir rahi hai."
$ρ$ = jahan vehicle hai wahan air density, kg/m³ mein (thicker air → zyada molecules takraenge → zyada heat).
$R_{n}$ = nose radius metres mein: nose jis ball jaisa dikhta hai uska radius. Bada ball = zyada gentle.
$v$ = air ke relative speed, m/s mein.
$k \approx 1.7 \times 1 0^{- 4}$ — ek empirical number jo formula ko wind-tunnel data se match karata hai. $ρ / R_{n}$ par square-root aur $v$ par cube kyun? Speed teen baar enter hoti hai (energy per mass $\sim v^{2}$ , aur wall se takrane wale mass ki rate $\sim v$ ), jisse $v^{3}$ milta hai. on $ρ / R_{n}$ woh boundary-layer scaling hai jo parent ne derive ki thi.

Definition

$k$ ki hidden units (taaki answer sach mein W/m² mein aaye) Bracketed pieces units carry karte hain: $ρ / R_{n}$ ki units hain $(kg m^{- 3}) / m = kg^{1/2} m^{- 2}$ , aur $v^{3}$ ki units hain $m^{3} s^{- 3}$ . In dono ko multiply karne par milta hai $kg^{1/2} m s^{- 3}$ . Ise $W/m^{2} = kg s^{- 3}$ mein convert karne ke liye, constant $k$ khud units carry karta hai $[k] = \frac{kg s ^{- 3}}{kg ^{1/2} m s ^{- 3}} = kg^{1/2} m^{- 1} .$ Toh $k$ ke "SI units" ka matlab hai $k = 1.7 \times 1 0^{- 4} kg^{1/2} m^{- 1}$ . Yahi reason hai ki $k$ ek empirical constant hai: yeh silently woh half-power of mass aur inverse metre absorb kar leta hai jo physics scaling khud fix nahi kar sakti.

Tool 2 — Re-radiation (Stefan–Boltzmann). Ek garam surface heat wapas bahar phenk deti hai: $\overset{q}{˙}_{rad,out} = ε σ T_{w}^{4}$

$ε$ = emissivity, ek pure number $0$ se (shiny, poorly radiate karta hai) lekar $1$ tak (perfect black radiator).
$σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ = Stefan–Boltzmann constant.
$T_{w}$ = wall temperature kelvin (K) mein. Dekho Radiative Heat Transfer and Stefan–Boltzmann Law.

Tool 3 — Ablator energy balance. Jo aata hai, minus jo re-radiate hota hai, woh material khaane mein jaata hai: $\overset{s}{˙} = \frac{q ˙ _{conv} - ε σ T _{w}^{4}}{ρ _{abl} Q ^{*}}$

$\overset{s}{˙}$ = recession rate: surface kitni tezi se andar khaati jaati hai, m/s mein.
$ρ_{abl}$ = ablator ki density (kg/m³).
$Q^{*}$ = effective heat of ablation (J/kg) — ek kilogram shield hatane mein kitni energy lagti hai.

Intuition Jab Tool 1 khud break ho jaata hai (radiation-dominated entry)

Sutton–Graves (Tool 1) ek convective correlation hai: yeh assume karta hai ki heat wall tak garm gas ke touch karne se pahunchi. Extreme speeds par (roughly $v > 10 km/s$ , Mach numbers ~25 ke baad — lunar aur Mars return) shock gas itna garam ho jaata hai ki woh glow karta hai, aur ek doosra channel khulta hai: gas heat seedha wall par radiate karta hai. Woh radiative input $\overset{q}{˙}_{rad,in}$ Tool 1 mein bilkul nahi hai, aur yeh roughly $ρ^{a} R_{n}^{b} v^{c}$ ke saath scale hota hai jisme $v$ ka power bahut steep hota hai (often $c \approx 8$ – $12$ , convective $v^{3}$ se kaafi upar). Toh bahut high speed par, convective correlations real load ko under-predict karti hain aur tumhe radiative term add karna padta hai — yeh exactly woh regime hai jiske edge par Ex 4 hai, aur isliye PICA (ek high- $Q^{*}$ ablator) wahan mandatory hai. Dekho Radiative Heat Transfer and Stefan–Boltzmann Law aur Reentry Aerothermodynamics.

Scenario matrix

Is page ka har problem neeche diye table ka ek cell hai. Agar koi cell cover nahi hua, toh ek din tum use unprepared milte — isliye hum sab cover kar rahe hain.

#	Cell class	Kya extreme / tricky hai	Example
C1	Baseline convective heating	ordinary LEO capsule	Ex 1
C2	Chota $R_{n}$ limit (sharp nose)	$R_{n} \to$ small → $\overset{q}{˙} \to$ huge	Ex 2
C3	Bada $R_{n}$ / density-zero limit	$R_{n} \to \infty$ ya $ρ \to 0$ → $\overset{q}{˙} \to 0$	Ex 3
C4	Speed scaling ( $v^{3}$ sensitivity, radiation onset)	$v$ double → ×8 heat; $v > 10$ km/s Tool 1 break karta hai	Ex 4
C5	Radiative equilibrium wall temp	tile survival, $T^{4}$ invert karo	Ex 5
C6	Ablator recession (saare terms live)	re-radiation subtract karo, $\overset{s}{˙}$ nikalo	Ex 6
C7	Degenerate ablation ( $\overset{s}{˙} \to 0$ )	re-radiation ≥ incoming → koi recession nahi	Ex 7
C8	Real-world word problem (mass budget)	Mars entry ke liye total shield mass	Ex 8
C9	Exam twist (kaun sa TPS + kyun)	$T_{w}$ melting se zyada → ablator force karta hai	Ex 9

Poore page mein use hone wale constants (inhe yaad karo): $k = 1.7 \times 1 0^{- 4} kg^{1/2} m^{- 1}, σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4} .$

Figure — Thermal protection systems — ablators (PICA, SLA), metallic tiles, RCC

Worked examples

Worked example Ex 1 — C1 Baseline: ek LEO capsule

Ek capsule $v = 7.5 km/s$ par reenter karta hai jahan air density $ρ = 1.0 \times 1 0^{- 3} kg/m^{3}$ hai. Nose radius $R_{n} = 1.0 m$ , $k = 1.7 \times 1 0^{- 4}$ . $\overset{q}{˙}_{conv}$ nikalo.

Forecast: Pehle order of magnitude guess karo — millions of watts per square metre? Tens of thousands? Aage padhne se pehle apna guess likho.

Speed ko SI mein convert karo: $v = 7500 m/s$ . Yeh step kyun? Formula SI mein hai; km/s silently answer ko $100 0^{3}$ se multiply kar deta.
Density/radius factor compute karo: $ρ / R_{n} = 1 0^{- 3} /1.0 = 1 0^{- 3} = 0.03162$ . Yeh step kyun? Yeh "kitne molecules, kitne spread out" wala part hai — ise isolate karo taaki $v$ ka cube alag rahe.
Speed cube karo: $v^{3} = (7500)^{3} = 4.219 \times 1 0^{11}$ . Yeh step kyun? Speed sabse dominant lever hai; ise sahi karna sabse zyada important hai.
Multiply karo: $\overset{q}{˙}_{conv} = 1.7 \times 1 0^{- 4} \times 0.03162 \times 4.219 \times 1 0^{11} = 2.27 \times 1 0^{6} W/m^{2}$ .

Answer: $\overset{q}{˙}_{conv} \approx 2.27 MW/m^{2}$ .

Verify: Units: $[k] \cdot kg m^{- 3} / m \cdot (m/s)^{3} = kg^{1/2} m^{- 1} \cdot kg^{1/2} m^{- 2} \cdot m^{3} s^{- 3} = kg s^{- 3} = W/m^{2}$ . ✓ Sanity: kuch $MW/m^{2}$ LEO return ke liye textbook figure hai — dekho Reentry Aerothermodynamics. Tumhara "millions" ka forecast sahi tha.

Worked example Ex 2 — C2 Sharp-nose limit: capsules round kyun hote hain

Ex 1 lo lekin nose ko $R_{n} = 0.01 m$ (1 cm sharp edge, jaise Shuttle wing leading edge) tak shrink karo. Same $ρ$ , same $v$ . Naya $\overset{q}{˙}$ aur Ex 1 se ratio nikalo.

Forecast: $R_{n}$ ×100 gira. Kyunki $\overset{q}{˙} \propto 1/ R_{n}$ , compute karne se pehle heating multiplier guess karo.

Naya factor: $ρ / R_{n} = 1 0^{- 3} /0.01 = 0.1 = 0.3162$ . Yeh step kyun? Sirf $R_{n}$ badla; Ex 1 ki baaki sab cheez reusable hai.
Ex 1 se ratio: $1.0/0.01 = 100 = 10$ . Yeh step kyun? $\overset{q}{˙} \propto 1/ R_{n}$ , toh ×100 chota radius $100 = 10$ × heat deta hai — yeh limit dekhne layak hai.
Naya heating: $\overset{q}{˙} = 10 \times 2.27 MW/m^{2} = 2.27 \times 1 0^{7} W/m^{2} = 22.7 MW/m^{2}$ .

Answer: $22.7 MW/m^{2}$ — blunt case ka das guna.

Verify: $R_{n} \to 0$ push karo (truly sharp): $1/ R_{n} \to \infty$ , toh $\overset{q}{˙} \to \infty$ . Yeh mathematical proof hai ki razor nose roast ho jaata hai — exactly isliye leading edges ko RCC chahiye, silica tiles nahi.

Worked example Ex 3 — C3 Vanishing limits: high altitude aur bada shield

Do degenerate checks. (a) Vehicle abhi bhi near-vacuum mein hai, $ρ = 1.0 \times 1 0^{- 8} kg/m^{3}$ ( $R_{n} = 1$ m, $v = 7500$ m/s). (b) Ek giant inflatable shield, $R_{n} = 100 m$ (wapas $ρ = 1 0^{- 3}$ ).

Forecast: Dono mein heating choti honi chahiye. $1 0^{5}$ ke factor ke change ke liye — density zyada tezi se giregi ya radius?

(a) Density factor: $1 0^{- 8} /1 = 1 0^{- 4}$ . Phir $\overset{q}{˙} = 1.7 \times 1 0^{- 4} \times 1 0^{- 4} \times 4.219 \times 1 0^{11} = 7.17 \times 1 0^{3} W/m^{2}$ . Yeh step kyun? Top-of-atmosphere limit dikhata hai: almost koi molecule nahi, almost koi heat nahi. Jaise $ρ \to 0$ , $\overset{q}{˙} \to 0$ smoothly.
(b) Radius factor: $1 0^{- 3} /100 = 1 0^{- 5} = 3.162 \times 1 0^{- 3}$ . Phir $\overset{q}{˙} = 1.7 \times 1 0^{- 4} \times 3.162 \times 1 0^{- 3} \times 4.219 \times 1 0^{11} = 2.27 \times 1 0^{5} W/m^{2}$ . Yeh step kyun? Jaise $R_{n} \to \infty$ , $\overset{q}{˙} \to 0$ — mathematical reason "blunter is cooler."

Answer: (a) $\approx 7.2 kW/m^{2}$ ; (b) $\approx 0.227 MW/m^{2}$ .

Verify: Density root ke neeche linearly enter hoti hai ( $ρ$ ); $ρ$ mein $1 0^{- 5}$ ki drop $1 0^{- 5} \approx 3 \times 1 0^{- 3}$ deti hai — same factor jo $1/ R_{n}$ mein $1 0^{- 5}$ ki drop deta. Toh symmetry hold karti hai: jab bhi koi variable gentle extreme ki taraf jaata hai, dono channels shut off ho jaate hain. Case (a) case (b) se $1000 \times$ chota hai kyunki $ρ$ $1 0^{5}$ gira jabki $R_{n}$ sirf $1 0^{2}$ badha.

Worked example Ex 4 — C4 Speed scaling:

$v^{3}$ lever (aur jahan Tool 1 break hota hai) Ek lunar-return capsule LEO ke $7.5 km/s$ ki jagah $v = 11 km/s$ hit karta hai (same $ρ = 1 0^{- 3}$ , $R_{n} = 1$ ). Convective heating kitne factor se jump karta hai, aur uski value kya hai? Phir caveat note karo.

Forecast: Speed $11/7.5 \approx 1.47$ badhi. $v^{3}$ ki wajah se, heat multiplier guess karo.

Speed ratio: $11/7.5 = 1.4667$ . Yeh step kyun? Sirf woh isolate karo jo badla.
Cube karo: $1.466 7^{3} = 3.155$ . Yeh step kyun? $\overset{q}{˙} \propto v^{3}$ , toh heating speed ratio ke cube ke saath scale hoti hai — reentry mein sabse important sensitivity.
Naya convective heating: $\overset{q}{˙}_{conv} = 3.155 \times 2.27 MW/m^{2} = 7.16 MW/m^{2}$ .

Answer: $\approx 3.16 \times$ zyada, $\approx 7.16 MW/m^{2}$ — sirf convective.

Verify: Sirf 47% tezi se triple heat ho gai. Caveat (radiation edge case): $11 km/s$ par hum $10 km/s$ ke baad hain, toh glowing-gas radiative input $\overset{q}{˙}_{rad,in}$ jo Tool 1 omit karta hai ab matter karta hai aur convective value ke barabar ya zyada ho sakta hai. $7.16 MW/m^{2}$ isliye ek lower bound hai; actual load zyada hai, aur exactly isliye deep-space returns (dekho Specific Impulse and Energy Budgets) high- $Q^{*}$ PICA ablator maangti hain. Direct check: $(11000)^{3} / (7500)^{3} = 3.155$ . ✓

Worked example Ex 5 — C5 Radiative equilibrium: kya silica tile survive karegi?

Ek reusable silica tile Shuttle belly par $\overset{q}{˙}_{conv} = 3.0 \times 1 0^{5} W/m^{2}$ face karti hai. Uski coating ki emissivity $ε = 0.85$ hai. Agar saari incoming heat re-radiate ho, toh steady wall temperature $T_{w}$ nikalo. (Tiles ~ $1500 K$ ke upar fail hoti hain.)

Forecast: Balance kehta hai $\overset{q}{˙}_{conv} = ε σ T_{w}^{4}$ . Hum fourth power invert karenge — kya $T_{w}$ 1500 K ke neeche ya upar aayega?

Balance likho: incoming = outgoing radiation, $\overset{q}{˙}_{conv} = ε σ T_{w}^{4}$ . Yeh step kyun? "Steady" ka matlab hai tile tab heating band karti hai jab jo radiate hota hai woh arrive hone ke barabar ho — energy in = energy out.
$T_{w}$ ke liye solve karo: $T_{w} = (\frac{q ˙ _{conv}}{ε σ})^{1/4}$ . Fourth root kyun? Radiation $T^{4}$ ke saath badhta hai; undo karne ke liye fourth root lete hain — inverse operation, bilkul jaise square undo karta hai.
Plug in karo: $\frac{3.0 \times 1 0 ^{5}}{0.85 \times 5.67 \times 1 0 ^{- 8}} = 6.226 \times 1 0^{12}$ . Fourth root: $T_{w} = (6.226 \times 1 0^{12})^{1/4} = 1580 K$ . Yeh step kyun? Inverted formula mein numbers substitute karo aur fourth root actually lo — yeh abstract balance ko ek concrete kelvin value mein convert karta hai jo hum 1500 K survival limit se compare kar saken.

Answer: $T_{w} \approx 1580 K$ .

Verify: 1500 K limit se thoda upar — yeh belly location tile ke survival edge ke kaafi paas hai, jo exactly real design margin ki kahani hai. Fourth root par sanity: $ε$ ko 0.85 se 1 ki taraf badhao toh $T_{w}$ girta hai (better radiator cooler chalta hai), jo high- $ε$ black coating choice confirm karta hai.

Worked example Ex 6 — C6 Full ablator: re-radiation live ke saath recession

Ek PICA shield ( $ρ_{abl} = 270 kg/m^{3}$ , $Q^{*} = 1.2 \times 1 0^{7} J/kg$ , $ε = 0.9$ ) char temperature $T_{w} = 3000 K$ par $\overset{q}{˙}_{conv} = 8.0 \times 1 0^{6} W/m^{2}$ ke neeche hai. Recession rate $\overset{s}{˙}$ nikalo.

Forecast: Pehle kuch heat re-radiate hoti hai; baaki material khaata hai. Guess karo ki 8 MW/m² mein se re-radiation bada ya chota slice hai.

Re-radiated flux: $ε σ T_{w}^{4} = 0.9 \times 5.67 \times 1 0^{- 8} \times (3000)^{4}$ . Kyunki $300 0^{4} = 8.1 \times 1 0^{13}$ , yeh hai $4.135 \times 1 0^{6} W/m^{2}$ . Yeh step kyun? Char white-hot glow karta hai; incoming heat ka bada chunk light ke roop mein nikal jaata hai, mass ablate hone se pehle hi.
Ablation ke liye net: $\overset{q}{˙}_{conv} - ε σ T_{w}^{4} = 8.0 \times 1 0^{6} - 4.135 \times 1 0^{6} = 3.865 \times 1 0^{6} W/m^{2}$ . Yeh step kyun? Sirf woh heat jo re-radiate nahi hoti, woh dying material ke zariye carry-off hoti hai.
Recession: $\overset{s}{˙} = \frac{3.865 \times 1 0 ^{6}}{270 \times 1.2 \times 1 0 ^{7}} = \frac{3.865 \times 1 0 ^{6}}{3.24 \times 1 0 ^{9}} = 1.193 \times 1 0^{- 3} m/s$ . Yeh step kyun? Net heat ko (shield ki energy per m³) $= ρ_{abl} Q^{*}$ se divide karo taaki metres per second mein recession mile.

Answer: $\overset{s}{˙} \approx 1.19 \times 1 0^{- 3} m/s \approx 1.19 mm/s$ .

Verify: Re-radiation ne incoming heat ka $4.1/8.0 \approx 52%$ khaya — aadhe se zyada — isliye hot char ek genuine third defence hai, footnote nahi. Units: $(W/m^{2}) / [(kg/m^{3}) (J/kg)] = (J s^{- 1} m^{- 2}) / (J m^{- 3}) = m/s$ . ✓

Worked example Ex 7 — C7 Degenerate ablation: recession zero par ruk jaata hai

Same PICA jaise Ex 6, lekin reentry ke end mein heating $\overset{q}{˙}_{conv} = 4.0 \times 1 0^{6} W/m^{2}$ par drop ho gayi hai jabki char abhi bhi $T_{w} = 3000 K$ par glow kar raha hai. $\overset{s}{˙}$ nikalo.

Forecast: Ex 6 se re-radiation term $4.135 MW/m^{2}$ hai. Incoming $4.0 MW/m^{2}$ hai. Kya hota hai jab shield zyada radiate kare jितना arrive ho raha hai?

Ablation ke liye net: $4.0 \times 1 0^{6} - 4.135 \times 1 0^{6} = - 1.35 \times 1 0^{5} W/m^{2}$ (negative). Yeh step kyun? Tool 3 mein bracket negative ja sakta hai — yeh woh degenerate case hai jo parent ne kabhi nahi dikhaya.
Negative ko interpret karo: negative "recession" unphysical hai — mass un-ablate nahi ho sakta. Physical outcome hai $\overset{s}{˙} = 0$ ; surface receding band ho jaati hai aur cool hone lagti hai (actual $T_{w}$ girta hai jab tak radiation reduced input ko balance na kare). Yeh step kyun? Hum zero par clamp karte hain kyunki formula ne steady ablation assume ki thi, jo tab valid nahi hoti jab re-radiation alone saari incoming heat dump kar sake.

Answer: $\overset{s}{˙} = 0$ (ablation ruk jaati hai; shield ab re-radiating tile ki tarah behave karta hai).

Verify: Yeh $\overset{q}{˙}_{conv} \leq ε σ T_{w}^{4}$ ka physical meaning hai: us threshold ke neeche koi material consume nahi hota. Yeh parent ka steel-man mistake "reusable tiles ablators se hamesha better hain" bhi reinforce karta hai — parent ka fix tha ki high flux par koi bhi reusable solid survive nahi kar sakta, isliye ablation required hai; yeh example flip side dikhata hai, ki low flux par ablator bhi ablate karna band kar deta hai aur reusable-jaisa behave karta hai. Dono milke prove karte hain ki TPS choice flux peak se govern hoti hai, average se nahi.

Worked example Ex 8 — C8 Real-world word problem: Mars shield mass

Ek Mars entry vehicle (dekho Mars Entry Descent and Landing) SLA-561V shield use karta hai, $ρ_{abl} = 260 kg/m^{3}$ . Engineers ko $4.5 m$ diameter ke circular heatshield par thickness $5.0 cm$ chahiye. Ablator ki mass kitni hogi?

Forecast: 4.5 m disc, 5 cm thick, roughly quarter-water density par. Guess karo: tens of kg? Hundreds?

Shield frontal area: $A = π r^{2} = π (2.25)^{2} = 15.90 m^{2}$ . Yeh step kyun? Ablator front face par ek layer hai; uska volume = area × thickness.
Volume: $V = A \times t = 15.90 \times 0.05 = 0.7952 m^{3}$ . Yeh step kyun? 5 cm ko pehle $0.05$ m mein convert karo taaki saari lengths SI mein rahein.
Mass: $m = ρ_{abl} V = 260 \times 0.7952 = 206.8 kg$ . Yeh step kyun? Mass = density × volume — density ki definition rearranged.

Answer: $\approx 207 kg$ ablator.

Verify: Kuch tonnes ke lander ke liye, ~200 kg heatshield ek believable single-digit-percent mass fraction hai — real EDL budgets se match karta hai. Units: $(kg/m^{3}) (m^{2}) (m) = kg$ . ✓

Worked example Ex 9 — C9 Exam twist: kaun sa TPS, aur numbers se prove karo

Ek wing leading edge $\overset{q}{˙}_{conv} = 1.5 \times 1 0^{6} W/m^{2}$ dekh raha hai. Ek candidate reusable ceramic $ε = 0.8$ ke saath radiate karta hai lekin 2000 K par melt ho jaata hai. Pure re-radiation use karte hue, decide karo ki ceramic survive karega ya nahi; agar nahi, toh TPS class ka naam batao jo ki isski jagah use hogi aur kyun.

Forecast: Radiative-equilibrium $T_{w}$ compute karo aur 2000 K se compare karo. Upar → material melt ho jaata hai.

Equilibrium temp: $T_{w} = (\frac{q ˙ _{conv}}{ε σ})^{1/4} = (\frac{1.5 \times 1 0 ^{6}}{0.8 \times 5.67 \times 1 0 ^{- 8}})^{1/4}$ . Yeh step kyun? Same balance jaise Ex 5 — ek reusable material tab survive kar sakta hai jab uska equilibrium temperature uske melting point ke neeche rahe.
Bracket ke andar: $\frac{1.5 \times 1 0 ^{6}}{4.536 \times 1 0 ^{- 8}} = 3.307 \times 1 0^{13}$ . Fourth root: $T_{w} = 2398 K$ . Yeh step kyun? $T^{4}$ ko fourth root se invert karo, phir numbers substitute karke concrete kelvin value lo jise 2000 K melting point se compare kar sakein.
Compare karo: $2398 K > 2000 K$ → ceramic melt ho jaata hai. Isliye ek reusable radiator is spot par survive nahi kar sakta. Ya toh RCC (carbon–carbon, >1900 K rated — Shuttle nose/leading-edge material) use karo ya ek ablator, jiska $\overset{m}{˙} Q^{*}$ blocking term woh energy remove karta hai jo koi solid re-radiate nahi kar sakta. Yeh step kyun? Yeh parent ka steel-man mistake quantitative banata hai: high flux par, $(\overset{q}{˙} / ε σ)^{1/4}$ har melting point se zyada ho sakta hai, jo ablation ya highest-temperature reusable material force karta hai.

Answer: $T_{w} \approx 2398 K$ , 2000 K se upar → ceramic fail; RCC ya ablator choose karo.

Verify: Yeh exactly Space Shuttle Columbia Accident ka lesson hai — leading edges RCC lete hain precisely isliye kyunki ordinary tiles ka equilibrium temperature unki limit se zyada ho jaata hai. 2398 K > 2000 K number bina kisi hand-waving ke decision settle kar deta hai.

Recall Matrix ko self-test karo

Kaun se cell mein fourth-root chahiye? ::: C5 aur C9 — $ε σ T_{w}^{4}$ ko invert karke $T_{w}$ nikalna. Kaun sa cell $\overset{q}{˙} \to \infty$ karta hai? ::: C2, jaise $R_{n} \to 0$ (sharp nose). Kaun sa cell $\overset{s}{˙}$ ko zero par clamp karta hai? ::: C7, jab re-radiation ≥ incoming heat ho. Speed double karne par heating kitne se multiply hoti hai? ::: $2^{3} = 8$ ( $v^{3}$ law, cell C4). Kis speed ke upar Tool 1 akela enough nahi rehta? ::: ~ $10 km/s$ (Mach >25), jahan glowing-gas radiative input oopar se add ho jaata hai.