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

3.4.22 · D2 · Physics › Rocket Flight Mechanics › Thermal protection systems — ablators (PICA, SLA), metallic

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Step 1 — "Surface par heat aana" ka matlab kya hai?

KYA. Kisi bhi formula se pehle, hume ek clean idea chahiye: heat flux. Ek chhoti si shield ki khidki imagine karo, exactly ek square metre, hot gas ki taraf munh karke. Ek second mein kuch amount of energy (joules mein) us khidki se shield ke andar jaati hai.

KYUN. Baad ke har symbol ya toh "energy per second per square metre" hai ya koi cheez jo ussi mein convert hoti hai. Agar pehle yeh unit picture clear na ho, toh balance equation sirf letters hai. Hum cheezein per square metre measure karte hain kyunki ek badi shield aur ek chhoti shield — same material ki bani hui — locally same behave karti hain — sirf local flux matter karta hai.

PICTURE. Figure dekho. Red square hamari ek-square-metre ki khidki hai. Black arrows andar aati energy hain. Hum energy-per-second-per-square-metre ko heat flux kehte hain aur ise $\dot{q}$ likhte hain.

• $\dot{q}$ ::: wall se takraane wali energy per second per square metre, ${\text{W/m}}^2$ mein
• "Per square metre" kyun? ::: taaki result is par depend na kare ki shield kitni badi hai — sirf local conditions par

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Step 2 — Yeh flux aata kahan se hai? (incoming term)

KYA. Reentry ke dauran wall par aane wala flux shock-heated gas se set hota hai. Reentry Aerothermodynamics aur parent note se, iska convective part scale karta hai: ${\dot{q}}_{\text{conv}}=k\sqrt{\frac{\rho }{R_n}}{v}^3.$

KYUN. Hum yeh isliye list karte hain taaki reader samjhe ki $\dot{q}$ koi free number nahi hai — yeh bahut bada hai (megawatts per square metre) aur yeh flight conditions se diya hua hai, shield se nahi. Shield ka kaam hai ki jo bhi yeh deliver kare, use survive kare. Hum wall mein enter karne ki koshish karne wale net heat ko ${\dot{q}}_{\text{net}}$ kahenge; ise socho "woh bacha hua flux jo abhi bhi andar ki taraf point kar raha hai, yeh account karne ke baad ki kya bahar ja raha hai."

PICTURE. Red arrow incoming ${\dot{q}}_{\text{net}}$ hai jo wall se takra raha hai. Formula ke neeche har symbol ko uski role ke saath label kiya gaya hai.

• $v^3$ kyun, $v^2$ kyun nahi? ::: energy per unit mass $v^2$ se scale karti hai, aur mass aane ki dar $v$ se scale karti hai; multiply karein → $v^3$
• Bada nose radius heating ke saath kya karta hai? ::: ise kam karta hai, kyunki $\dot{q}\propto 1/\sqrt{R_n}$

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Step 3 — Wall wapas ladhti hai: yeh re-radiate karti hai

KYA. Ek hot surface wahan bas baithti nahi — yeh glow karti hai. Temperature $T_w$ ("w" wall ke liye hai) par koi bhi object energy wapas bahar throw karta hai light aur infrared ke roop mein. Amount hai: ${\dot{q}}_{\text{out}}=\epsilon \sigma T_w^4.$

KYUN. Hum yeh isliye introduce karte hain kyunki sabhi incoming flux shield material ko destroy karne mein nahi lagti — kuch wapas radiation ke roop mein nikl jaati hai. Agar hum yeh term bhool jaate toh hum bahut bura overestimate karte ki shield kitni tezi se erode hoti hai. Yeh Stefan–Boltzmann law hai, Radiative Heat Transfer and Stefan–Boltzmann Law se borrowed.

PICTURE. Wohi square, lekin ab black arrows surface se nikal rahe hain. Wall jitni zyada hot hogi, utni zyada glow karegi — aur fourth power ki wajah se, temperature double karne par sixteen guna zyada radiate hoga.

• $T_w$ kelvin mein kyun hona chahiye? ::: radiation absolute temperature par depend karta hai; $T^4$ sirf absolute zero se meaningful hai
• Char ko black (high $\epsilon$) kyun paint karte hain? ::: high $\epsilon$ matlab yeh zyada radiate karta hai, toh structure tak pahunchne se pehle zyada heat nikal jaati hai

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Step 4 — Jo bacha raha woh dying material ko swallow karna padta hai

KYA. Incoming flux lo, jo radiate ho gaya usse subtract karo. Jo bhi bacha hai woh energy hai jo physically shield material ko destroy karne mein jaati hai — chemical bonds todna, melting, boiling. Per second per square metre remove ki gayi mass ko $\dot{m}$ kaho, aur ek kilogram remove karne ke liye chahiye energy ko $Q^{\ast }$. Tab per second per square metre ablation jo energy soakta hai woh hai $\dot{m}{Q}^{\ast }$.

KYUN. Yeh ablator idea ka dil hai, "survive karne ke liye maro." Surface par energy conserve honi chahiye: andar aane wali sabhi cheez ki accounting honi chahiye (light ke roop mein nikalna) + (material marne mein kharchna) se. Steady state mein iske jaane ki aur koi jagah nahi hai.

PICTURE. Red layer shield material hai jo gas ban kar stream away ho rahi hai; chhote arrows bahar jaati mass hain. Jo energy woh carry off karti hai woh hai $\dot{m}{Q}^{\ast }$.

• $Q^{\ast }$ ka matlab ::: destroy ki gayi shield ke per kilogram absorb ki gayi energy
• High $Q^{\ast }$ desirable kyun hai? ::: shield ka har kilogram zyada heat swallow karta hai, toh same heat load ke liye kam material khota hai

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Step 5 — Balance likho aur ise ek equation ki tarah padho

KYA. Teeno flows ko ek line par rakho. Andar aata hai ${\dot{q}}_{\text{net}}$; bahar jaata hai radiation $\epsilon \sigma T_w^4$; baaki ablation $\dot{m}{Q}^{\ast }$ hai: ${\dot{q}}_{\text{net}}=\dot{m}{Q}^{\ast }+\epsilon \sigma T_w^4.$

KYUN. Yeh sirf "books balance" hai: energy in = energy accounted for. Koi naya physics nahi — sirf teeno arrows ki bookkeeping jo humne Steps 2, 3, 4 mein draw ki thi. Yeh single line parent note ki boxed energy balance hai.

PICTURE. Ek square par teeno arrows: red = incoming ${\dot{q}}_{\text{net}}$, black upar = radiation, black door = ablation mass. Equation yeh promise hai ki red dono blacks ke combined barabar hai.

• Yeh line kaunsa principle hai? ::: surface par conservation of energy — in equals out
• Agar wall cold ho toh kaunsa term gayab ho jaata hai? ::: radiation term $\epsilon \sigma T_w^4$, $T_w\to 0$ ke saath zero ki taraf shrink hota hai

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Step 6 — Woh solve karo jo ek engineer actually chahta hai: recession rate

KYA. Ek engineer directly $\dot{m}$ measure nahi karta — woh jaanna chahta hai surface kitni tezi se recede karti hai, yaani shield ke kitne millimetre per second gayab ho jaate hain. Use $\dot{s}$ kaho (s surface position ke liye). Mass removed aur thickness removed material ki density ${\rho }_{\text{abl}}$ se linked hain: $\dot{m}={\rho }_{\text{abl}}\dot{s}.$ Balance mein substitute karo aur $\dot{s}$ ke liye solve karo:

\;\Longrightarrow\; \boxed{\;\dot s=\frac{\dot q_\text{net}-\varepsilon\sigma T_w^4}{\rho_\text{abl}\,Q^*}\;}$$ **KYUN.** Hum $\rho_\text{abl}Q^*$ se divide karte hain $\dot s$ isolate karne ke liye kyunki woh design number hai: $\dot s$ ko flight time se multiply karo aur tumhe pata chalega shield kitni thick honi chahiye taaki woh poori jal na jaye. Numerator mein subtraction keh rahi hai ki wall tabhi mass khona shuru karti hai jab flux *re-radiation ke baad bacha* ho. **PICTURE.** Red line retreating surface hai; arrow $\dot s$ dikha raha hai ki yeh shield mein kuch millimetre per second ki dar se peeche ki taraf ja rahi hai. Purani surface (dashed) ja chuki hai. > [!formula] Recession rate — target result > $$\dot s=\frac{\dot q_\text{net}-\varepsilon\sigma T_w^4}{\rho_\text{abl}\,Q^*}$$ > - Numerator $\dot q_\text{net}-\varepsilon\sigma T_w^4$ — radiation ke baad bacha *net* flux; sirf yahi material loss drive karta hai. > - $\rho_\text{abl}$ — shield density; ek lighter, low-density material (PICA at $270\ \text{kg/m}^3$) per kilogram *tezi se* recede karta hai lekin overall kam weighs karta hai. > - $Q^*$ — high value = slow recession. ![[deepdives/dd-physics-3.4.22-d2-s06.png]] - $\rho_\text{abl}Q^*$ se divide kyun karte hain? ::: "bacha hua energy" ko "thickness lost per second" mein convert karne ke liye, woh quantity jisse ek designer shield size karta hai - Required shield thickness pane ke liye $\dot s$ se kya multiply karte hain? ::: total reentry duration (heating time) --- ## Step 7 — Edge cases: formula kahan strange behave karta hai? **KYA.** Ek accha formula apne extremes survive karna chahiye. Teen cases: 1. **Cold start ($T_w\approx0$).** Radiation term $\varepsilon\sigma T_w^4\to0$, toh $\dot s\to\dot q_\text{net}/(\rho_\text{abl}Q^*)$ — *maximum* recession. Shield pehle heating ke instant par sabse tezi se erode karti hai, pehle ki woh warm up hoke glow karna shuru kare. 2. **Radiation-dominated ($\varepsilon\sigma T_w^4=\dot q_\text{net}$).** Numerator zero ho jaata hai, toh $\dot s=0$ — **bilkul bhi ablation nahi**. Saari incoming heat light ke roop mein nikal jaati hai. Yeh parent note se exactly *reusable-tile* limit hai: ek tile tabhi survive karti hai jab woh sab kuch radiate away kar sake. Dekho [[Radiative Heat Transfer and Stefan–Boltzmann Law]]. 3. **Over-radiating? ($\varepsilon\sigma T_w^4>\dot q_\text{net}$).** Numerator negative ho jaata hai, $\dot s<0$. Negative recession unphysical hai — tum shield wapas ugra nahi sakte. Iska matlab actually hai: wall itni hot *nahi reh sakti*; yeh receive karne se zyada radiate karegi, toh yeh cool hogi jab tak balance wapas na aaye. Equation khud theek ho jaata hai $T_w$ ko neeche drive karke jab tak numerator $\ge0$ na ho. **KYUN.** Yeh cases is ek formula ko parent mein puri *TPS decision table* se connect karte hain: case 2 ek tile hai, cases 1 aur 3 ek real ablator ki zindagi bracket karte hain. Parent ki story mein kuch bhi is equation ke bahar nahi hai. **PICTURE.** Teen mini-panels: (left) cold, fast recession; (middle) balanced, zero recession = tile limit; (right) impossible negative case jisme ek arrow $T_w$ ko balance par wapas girta dikha raha hai. ![[deepdives/dd-physics-3.4.22-d2-s07.png]] > [!mistake] "Agar $T_w$ bahut bada ho toh shield negative recede karta hai — equation toot gayi." > **Kyun sahi lagta hai:** tum ek bada $T_w$ plug in kar sakte ho aur $\dot s<0$ pa sakte ho. > **Fix:** $T_w$ koi free knob nahi hai. Steady state mein wall temperature *settle* hoti hai jis bhi value par jo books ko $\dot s\ge0$ ke saath balance kare. Negative answer sirf batata hai ki tumhara assumed $T_w$ bahut zyada tha; physically yeh tab tak girta hai jab tak $\dot s=0$ (tile case) na ho ya jab tak ablation restart na ho. - Cold-wall limit kaisi recession deta hai? ::: maximum, $\dot s=\dot q_\text{net}/(\rho_\text{abl}Q^*)$, kyunki radiation negligible hai - Zero-recession condition ::: $\varepsilon\sigma T_w^4=\dot q_\text{net}$ — sab kuch re-radiate hota hai, reusable-tile limit --- ## Step 8 — Ek worked number taaki yeh real lage **KYA.** Ek moderate reentry point par PICA-jaisi shield lo: $\dot q_\text{net}=2.0\times10^{6}\ \text{W/m}^2$, wall glowing at $T_w=2000\ \text{K}$, $\varepsilon=0.9$, $\sigma=5.67\times10^{-8}$, density $\rho_\text{abl}=270\ \text{kg/m}^3$, aur $Q^*=1.2\times10^{7}\ \text{J/kg}$. **KYUN.** Numbers algebra ko "kya meri shield survive karegi" mein convert karte hain. Hum radiated flux, net driving flux, aur finally $\dot s$ compute karte hain. **Radiated flux:** $$\varepsilon\sigma T_w^4=0.9\times5.67\times10^{-8}\times(2000)^4=0.9\times5.67\times10^{-8}\times1.6\times10^{13}\approx8.16\times10^{5}\ \text{W/m}^2.$$ **Recession rate:** $$\dot s=\frac{2.0\times10^{6}-8.16\times10^{5}}{270\times1.2\times10^{7}}=\frac{1.184\times10^{6}}{3.24\times10^{9}}\approx3.65\times10^{-4}\ \text{m/s}.$$ Yeh lagbhag **0.37 mm per second** hai. 120-second ke heat pulse mein, shield lagbhag $0.37\times120\approx44\ \text{mm}$ khoti hai — toh ek 6 cm shield margin ke saath survive karti hai. Notice karo ki glowing wall ne incoming flux ka lagbhag **41\%** radiation se bahar shed kar diya, koi bhi material lose hone se pehle. ![[deepdives/dd-physics-3.4.22-d2-s08.png]] - Worked case ke liye recession rate ::: lagbhag $3.65\times10^{-4}$ m/s, yaani $\approx0.37$ mm/s - Incoming flux ka kitna fraction radiate ho gaya ::: $\approx0.816/2.0\approx41\%$ --- ## Ek-picture summary ![[deepdives/dd-physics-3.4.22-d2-s09.png]] Ek frame mein poori derivation: incoming red flux $\dot q_\text{net}$ wall se takrata hai; kuch glow $\varepsilon\sigma T_w^4$ ke roop mein nikalta hai; baaki surface se mass $\dot m\,Q^*$ drive karta hai; $\rho_\text{abl}$ se mass loss ko thickness loss mein convert karne par recession rate $\dot s$ milti hai. > [!recall]- Poore walkthrough ki Feynman retelling > Heat shield ke ek ek-square-metre patch ki picture banao. Reentry gas usse energy ka ek firehose dump karta hai — lakhon watts per square metre (Step 1–2). Patch do tareekon se ladta hai. Pehla, jab woh glowing hot ho jaata hai toh energy *wapas bahar* light ke roop mein throw karta hai, aur kyunki glow temperature ke fourth power ke saath badhta hai, ek bright char incoming heat ka ek bada chunk shed kar deta hai (Step 3). Doosra, jo bhi heat *abhi bhi bacha* hai woh shield material ko apart rip karta hai — bonds todta hai, ise boil off karta hai — aur woh gas door stream karta hai apne saath energy le jaata hai; uthaaya gaya har kilogram $Q^*$ joules khata hai (Step 4). Energy balance honi chahiye: jo aata hai woh equals karta hai jo glows out plus jo boils off (Step 5). Kyunki engineers ko iski parwah hai ki *shield kitni tezi se shrink hoti hai*, hum "mass per second" ko "millimetre per second" mein swap karte hain material ki density use karke aur solve karte hain — yahi recession rate $\dot s$ hai (Step 6). Extremes check karo: ek cold wall sabse tezi se erode hoti hai, ek wall jo sab kuch radiate kar sake bilkul erode nahi hoti (yahi reusable tile hai), aur agar tum assume karo ki wall bahut zyada hot hai toh math kehta hai "impossible," jiska matlab aslaan hai ki wall cool hoti hai jab tak books balance na ho jaayein (Step 7). Real PICA numbers plug in karo aur pata chalega ki shield aadhe millimetre per second se kam retreat kar rahi hai — kuch centimetre material poore fiery plunge se bacha rehta hai (Step 8). > [!mnemonic] > **IN = GLOW + GONE.** Incoming heat equals jo glow away hota hai plus jo gone (ablated) ho jaata hai. "Gone" ko density se divide karo pata karne ke liye ki surface kitni tezi se shrink hoti hai. --- **See also:** [[Bow Shock and Stagnation Point]] · [[Blunt Body Aerodynamics]] · [[Space Shuttle Columbia Accident]] · [[Mars Entry Descent and Landing]] · [[Specific Impulse and Energy Budgets]]