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

3.4.22 · D4 · Physics › Rocket Flight Mechanics › Thermal protection systems — ablators (PICA, SLA), metallic

Shuru karne se pehle, aao har ek symbol ko pin down karein jo hum baar baar use karenge, taaki kuch bhi unexplained na lage:

Definition Is page ke symbols

$v$ — vehicle ki speed, air ke relative mein, metres per second mein ( $m/s$ ). "Kitni tez se ye crash karta hai."
$ρ$ (Greek "rho") — aaspaas ki hawa ki density, kilograms per cubic metre mein ( $kg/m^{3}$ ). Patli, oopar wali hawa = chhota $ρ$ .
$R_{n}$ — nose radius: socho ki vehicle ke rounded front mein sab se badi ball fit kar rahe ho; uska radius $R_{n}$ hai, metres mein. Blunt = bada $R_{n}$ .
$\overset{q}{˙}$ (padho "q-dot") — ek heat flux: har second mein wall ke har square metre par pahunchne wali energy, watts per square metre mein ( $W/m^{2}$ ). Dot matlab "per second".
$σ$ (Greek "sigma") — Stefan–Boltzmann constant $= 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ . Ye set karta hai ki ek garam surface kitni strongly glow karti hai. (Dekho Radiative Heat Transfer and Stefan–Boltzmann Law.)
$ε$ (Greek "epsilon") — emissivity, ek pure number $0$ se $1$ tak. $ε = 1$ ek perfect radiator hai ("perfectly black"); $ε = 0$ kuch bhi radiate nahi karta.
$T_{w}$ — wall temperature kelvin mein ( $K$ ). Kelvin = Celsius $+ 273$ ; ye kabhi negative nahi jaata.
$Q^{*}$ — effective heat of ablation, shield ke har kilogram ke khaane mein absorbed energy ( $J/kg$ ). Bada $Q^{*}$ = har gram zyada protect karta hai.
$\overset{s}{˙}$ — recession rate: shield ki thickness ke kitne metres har second mein jal jaate hain ( $m/s$ ).
$ρ_{abl}$ — ablator material ki khud ki density ( $kg/m^{3}$ ).

Level 1 — Recognition

L1.1

In do TPS families mein se kaun si flight ke dauran consume hoti hai, aur kaun si reuse hoti hai: (a) PICA ablator, (b) silica tiles?

Recall Solution

HUM KYA KARTE HAIN: parent note se do philosophies yaad karte hain.

(a) PICA ek ablator hai → ye consumed hoti hai: ye decompose hoti hai, vaporise hoti hai, aur heat shedne ke liye gas bahar phenkti hai.
(b) Silica tiles reusable hain → ye $ε σ T_{w}^{4}$ re-radiate karke aur insulate karke survive karti hain, koi mass nahi khoti.

Memory hook: ablators marte hain bachne ke liye; tiles glow karke bachti hain.

L1.2

Teen simultaneous mechanisms batao jinse ek ablator heat shedta hai.

Recall Solution

Pyrolysis / phase change — bonds toot jaate hain aur material vaporise ho jaati hai, latent heat absorb karke.
Blowing (blockage / transpiration) — nikli hui gas hot boundary layer ko wall se door dhakelt hai.
Re-radiation — kala char surface glow karta hai aur $ε σ T_{w}^{4}$ bahar radiate karta hai.

Level 2 — Application

L2.1 — Reentry energy budget

Ek capsule low Earth orbit se $v = 7.8 km/s$ par return karta hai. Uski specific kinetic energy $e_{k} = \frac{1}{2} v^{2}$ calculate karo $MJ/kg$ mein.

Recall Solution

$\frac{1}{2} v^{2}$ kyun: mass per unit kinetic energy mass ko strip kar deti hai, sirf speed bacha kar — ye woh energy hai jo vehicle ke har kilogram mein destroy karni hoti hai. $e_{k} = \frac{1}{2} (7800)^{2} = \frac{1}{2} \times 6.084 \times 1 0^{7} = 3.04 \times 1 0^{7} J/kg \approx 30 MJ/kg .$ Comparison ke liye, ek kilogram steel ko boil karne mein ~ $7 MJ/kg$ lagta hai — reentry energy uska karib chaar guna hai. Isliye TPS exist karta hai. Dekho Specific Impulse and Energy Budgets.

L2.2 — Sutton–Graves heat flux

$\overset{q}{˙}_{conv} = k ρ / R_{n} v^{3}$ use karo, $k = 1.7 \times 1 0^{- 4}$ (SI), $ρ = 1.0 \times 1 0^{- 3} kg/m^{3}$ , $R_{n} = 0.5 m$ , $v = 7000 m/s$ ke saath, $\overset{q}{˙}_{conv}$ nikalo $MW/m^{2}$ mein.

Recall Solution

Ye formula kyun: stagnation heating scale hoti hai (energy per mass $\sim v^{2}$ ) × (mass flux $\sim ρ v$ ) × (boundary-layer factor $\sim 1/ R_{n}$ ) ke saath, jo deta hai $ρ v^{3} / R_{n}$ ; $ρ / R_{n}$ ka square root data-fitted form hai. $\frac{ρ}{R _{n}} = \frac{1 0 ^{- 3}}{0.5} = 2 \times 1 0^{- 3} = 0.04472.$ $v^{3} = (7000)^{3} = 3.43 \times 1 0^{11} .$ $\overset{q}{˙} = 1.7 \times 1 0^{- 4} \times 0.04472 \times 3.43 \times 1 0^{11} = 2.61 \times 1 0^{6} W/m^{2} \approx 2.6 MW/m^{2} .$

L2.3 — Radiative equilibrium temperature

Ek reusable tile saari incoming heat re-radiate karta hai: $\overset{q}{˙}_{conv} = ε σ T_{w}^{4}$ . $\overset{q}{˙}_{conv} = 0.30 MW/m^{2}$ , $ε = 0.85$ ke saath, $T_{w}$ nikalo.

Recall Solution

Fourth root kyun use karte hain: surface steady state tab reach karta hai jab jo wo radiate karta hai exactly match kare jo arrive karta hai. Radiation $T^{4}$ ke saath badhti hai, isliye temperature nikalne ke liye fourth power undo karte hain — fourth root lete hain. $T_{w} = (\frac{q ˙ _{conv}}{ε σ})^{1/4} = (\frac{3.0 \times 1 0 ^{5}}{0.85 \times 5.67 \times 1 0 ^{- 8}})^{1/4} .$ Andar: $\frac{3.0 \times 1 0 ^{5}}{4.82 \times 1 0 ^{- 8}} = 6.23 \times 1 0^{12}$ . Fourth root: $T_{w} = (6.23 \times 1 0^{12})^{1/4} \approx 1580 K (\approx 1300^{\circ} C) .$ Silica tile ki limit ke andar — isliye is modest flux par ek reusable tile kafi hai.

Neeche wali figure exactly yahi relationship plot karti hai: padho kaise equilibrium wall temperature badhti hai jaise incoming flux badhta hai. Plum dot dhundho $0.3 MW/m^{2}$ par — woh L2.3 hai jo safely orange silica-melt line ke neeche land kar raha hai — aur ink dot note karo jo bahut daayein hai, jise hum L4.1 mein milenge.

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

Level 3 — Analysis

L3.1 — Ablator ki recession rate

Ek ablator net wall heat $\overset{q}{˙}_{net} = 5.0 MW/m^{2}$ face karta hai, $ε σ T_{w}^{4}$ re-radiate karta hai jisme $ε = 0.9$ , $T_{w} = 2500 K$ hai. Material: $ρ_{abl} = 270 kg/m^{3}$ (PICA), $Q^{*} = 1.0 \times 1 0^{7} J/kg$ . Recession rate $\overset{s}{˙}$ nikalo $mm/s$ mein.

Recall Solution

Energy balance kya kehti hai: jo heat radiate nahi hoti use material ke khaane se uthana padta hai. Toh pehle re-radiated part subtract karo. $ε σ T_{w}^{4} = 0.9 \times 5.67 \times 1 0^{- 8} \times (2500)^{4} = 0.9 \times 5.67 \times 1 0^{- 8} \times 3.906 \times 1 0^{13} = 1.99 \times 1 0^{6} W/m^{2} .$ Net ablation mein: $5.0 \times 1 0^{6} - 1.99 \times 1 0^{6} = 3.01 \times 1 0^{6} W/m^{2}$ . $ρ_{abl} Q^{*}$ se kyun divide karte hain: shield ka har cubic metre $ρ_{abl} Q^{*}$ joules protection store karta hai; bacha hua flux isse divide karne par pata chalta hai ki thickness kitni tez ghatti hai. $\overset{s}{˙} = \frac{q ˙ _{net} - ε σ T _{w}^{4}}{ρ _{abl} Q ^{*}} = \frac{3.01 \times 1 0 ^{6}}{270 \times 1 0 ^{7}} = 1.11 \times 1 0^{- 3} m/s = 1.11 mm/s .$

L3.2 — Kyun blunt bodies thande rehte hain

Ek designer nose radius $R_{n} = 0.25 m$ se double karke $0.50 m$ kar deta hai, baaki sab fixed. $\overset{q}{˙}_{conv}$ kis factor se change hoga?

Recall Solution

Yahan sirf $R_{n}$ kyun matter karta hai: $\overset{q}{˙} \propto ρ / R_{n} v^{3}$ mein, $R_{n}$ double karne se $ρ$ aur $v$ wahi rehte hain, isliye sirf $1/ R_{n}$ factor change hota hai. $\frac{q ˙ _{new}}{q ˙ _{old}} = \frac{R _{n, old}}{R _{n, new}} = \frac{0.25}{0.50} = 0.5 = 0.707.$ Heating ~71% ho jaati hai — sirf nose ko aur round karne se ~29% reduction. Ye hai Blunt Body Aerodynamics ek line mein, aur isliye capsules round hote hain.

Neeche wali figure $1/ R_{n}$ curve draw karti hai. Isko plum dot (sharp leading edge, tiny $R_{n}$ , high heating) se teal dot (blunt belly, large $R_{n}$ , low heating) tak trace karo — left par steep drop exactly yahi explain karta hai ki sharp noses kyun jalte hain aur blunt ones kyun thande rehte hain.

Level 4 — Synthesis

L4.1 — High flux par Ablator vs tile

$\overset{q}{˙}_{conv} = 10 MW/m^{2}$ par, kya ek silica tile ( $ε = 0.85$ , pighalta hai ~ $1900 K$ ) purely re-radiating surface ki tarah survive kar sakti hai? Required equilibrium $T_{w}$ compute karo aur decide karo.

Recall Solution

HUM KYA COMPUTE KARTE HAIN: woh temperature jo ek purely radiating surface ko $10 MW/m^{2}$ shedne ke liye chahiye. $T_{w} = (\frac{1 0 ^{7}}{0.85 \times 5.67 \times 1 0 ^{- 8}})^{1/4} = (\frac{1 0 ^{7}}{4.82 \times 1 0 ^{- 8}})^{1/4} = (2.075 \times 1 0^{14})^{1/4} .$ $T_{w} \approx 3800 K .$ DECISION: $3800 K$ ~ $1900 K$ silica melting point se bahut zyada hai. Ek re-radiating tile survive nahi kar sakti — tumhe ek ablator use karna hoga, jiska $\overset{m}{˙} Q^{*}$ blocking term woh energy remove karta hai jo koi solid radiate nahi kar sakta. Ye quantitative heart hai "reusable always better nahi hota" ki baat ka.

L4.2 — Total shield thickness kitni chahiye

Ek deep-space capsule L3.1 conditions ( $\overset{s}{˙} = 1.11 mm/s$ , constant maano) $90 s$ ke heat pulse ke liye face karta hai. Minimum kitna ablator consume hoga, aur $50%$ safety margin add karo.

Recall Solution

Rate ko time se kyun multiply karte hain: agar thickness steady $\overset{s}{˙}$ se jalti hai, total loss bas $\overset{s}{˙} \times t$ hai (rate × time ka ek rectangle). $Δ s = \overset{s}{˙} t = 1.11 \times 1 0^{- 3} \times 90 = 0.0999 m \approx 100 mm .$ $50%$ margin ke saath: $100 \times 1.5 = 150 mm$ . Toh heat shield kam se kam 15 cm moti honi chahiye — real deep-space shields (jaise Stardust-class PICA) sach mein is order ki thickness ke hote hain.

Level 5 — Mastery

L5.1 — Mars return par convective vs radiative crossover

Fast Mars/lunar return par, radiative heating rivals convective heating ko. Maano convective heating $v^{3}$ se scale hoti hai aur radiative heating (roughly) $v^{8}$ se interest ke range mein. Agar $v_{1} = 7 km/s$ par $\overset{q}{˙}_{conv} = \overset{q}{˙}_{rad}$ equal hain, toh $v_{2} = 11 km/s$ par ratio $\overset{q}{˙}_{rad} / \overset{q}{˙}_{conv}$ kya hoga?

Recall Solution

Scalings ko kyun divide karte hain: kyunki $\overset{q}{˙}_{conv} \propto v^{3}$ aur $\overset{q}{˙}_{rad} \propto v^{8}$ , unka ratio $v^{8} / v^{3} = v^{5}$ se scale hota hai. Ye $v_{1}$ par equal se shuru hote hain, isliye: $\frac{q ˙ _{rad}}{q ˙ _{conv}}_{v_{2}} = (\frac{v _{2}}{v _{1}})^{5} = (\frac{11}{7})^{5} .$ $(\frac{11}{7})^{5} = (1.5714)^{5} \approx 9.57.$ Radiative heating convective se karib 10 guna badi ho jaati hai — exactly isliye deep-space return mein high- $Q^{*}$ ablators jaise PICA chahiye, tiles nahi, aur isliye radiative blockage matter karta hai. Dekho Mars Entry Descent and Landing.

L5.2 — RCC kahan jaata hai, aur kyun?

Shuttle wing leading edge ka ek chhota local nose radius $R_{n} = 0.03 m$ hai; belly tile region ka effective $R_{n} = 1.0 m$ hai. Same $ρ$ aur $v$ lo. Leading-edge heat flux ka belly heat flux se ratio nikalo, aur material choice explain karo.

Recall Solution

Sirf $R_{n}$ kyun enter karta hai: identical $ρ$ , $v$ ⇒ ratio sirf $1/ R_{n}$ se set hota hai. $\frac{q ˙ _{LE}}{q ˙ _{belly}} = \frac{R _{n, belly}}{R _{n, LE}} = \frac{1.0}{0.03} = 33.3 = 5.77.$ Leading edge belly ki ~5.8 guna heat flux dekhta hai. Silica tiles (~ $1900 K$ limit) ye nahi le sakti; isliye sharp, sabse garam spots RCC use karte hain (survive karta hai $> 1600^{\circ} C \approx 1900 K$ aur aage bhi). Space Shuttle Columbia Accident ek RCC leading-edge breach thi — proof ki yeh sabse garam zone kitna critical hai.

Recall Self-test summary

Har level ke liye ek-line why ::: L1 families jaanta hai; L2 $\frac{1}{2} v^{2}$ , Sutton–Graves, aur $T_{w} = (\overset{q}{˙} / ε σ)^{1/4}$ mein plug karta hai; L3 ablation energy aur $1/ R_{n}$ blunt-body law balance karta hai; L4 dikhata hai ki radiation ki ek hard temperature ceiling hai jo ablation force karta hai; L5 $v^{5}$ crossover aur per-zone material selection handle karta hai.