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

WHAT we do: recall the two philosophies from the parent note.

• (a) PICA is an ablator → it is consumed: it decomposes, vaporises, and blows gas away to shed heat.
• (b) Silica tiles are reusable → they survive by re-radiating $\epsilon \sigma T_w^4$ and insulating, losing no mass.

Memory hook: ablators die to survive; tiles survive by glowing.

1. Pyrolysis / phase change — bonds break and material vaporises, absorbing latent heat.
2. Blowing (blockage / transpiration) — the outgas pushes the hot boundary layer away from the wall.
3. Re-radiation — the black char surface glows and radiates $\epsilon \sigma T_w^4$ back out.

WHY $\frac{1}{2}{v}^2$: kinetic energy per unit mass strips out the mass, leaving only speed — this is the energy that must be destroyed per kilogram of vehicle. $e_k=\frac{1}{2}(7800{)}^2=\frac{1}{2}\times 6.084\times 10^7=3.04\times 10^7\text{J/kg}\approx 30\text{MJ/kg}.$ For comparison, boiling a kilogram of steel costs ~$7\text{MJ/kg}$ — the reentry energy is about four times that. This is why a TPS exists. See also Specific Impulse and Energy Budgets.

WHY this formula: stagnation heating scales as (energy per mass $\sim v^2$) × (mass flux $\sim \rho v$) × (boundary-layer factor $\sim 1/\sqrt{R_n}$), giving $\rho v^3/\sqrt{R_n}$; the square root of $\rho /R_n$ is the data-fitted form. $\sqrt{\frac{\rho }{R_n}}=\sqrt{\frac{10^{-3}}{0.5}}=\sqrt{2\times 10^{-3}}=\mathrm{0.04472.}$ $v^3=(7000{)}^3=3.43\times 10^{11}.$ $\dot{q}=1.7\times 10^{-4}\times 0.04472\times 3.43\times 10^{11}=2.61\times 10^6{\text{W/m}}^2\approx 2.6{\text{MW/m}}^2.$

WHY invert to a fourth root: the surface reaches steady state when what it radiates exactly matches what arrives. Radiation grows as $T^4$, so to find the temperature we undo the fourth power — take the fourth root. $T_w={\left(\frac{{\dot{q}}_{\text{conv}}}{\epsilon \sigma }\right)}^{1/4}={\left(\frac{3.0\times 10^5}{0.85\times 5.67\times 10^{-8}}\right)}^{1/4}.$ Inside: $\frac{3.0\times 10^5}{4.82\times 10^{-8}}=6.23\times 10^{12}$. Fourth root: $T_w=(6.23\times 10^{12}{)}^{1/4}\approx 1580\text{K}(\approx 1300{}^{\circ }\text{C}).$ Well within a silica tile's limit — so at this modest flux, a reusable tile suffices.

WHAT the energy balance says: whatever heat is not radiated away must be carried off by material being eaten. So first subtract the re-radiated part. $\epsilon \sigma T_w^4=0.9\times 5.67\times 10^{-8}\times (2500{)}^4=0.9\times 5.67\times 10^{-8}\times 3.906\times 10^{13}=1.99\times 10^6{\text{W/m}}^2.$ Net into ablation: $5.0\times 10^6-1.99\times 10^6=3.01\times 10^6{\text{W/m}}^2$. WHY divide by ${\rho }_{\text{abl}}{Q}^{\ast }$: each cubic metre of shield stores ${\rho }_{\text{abl}}{Q}^{\ast }$ joules of protection; dividing the leftover flux by it gives how fast thickness disappears. $\dot{s}=\frac{{\dot{q}}_{\text{net}}-\epsilon \sigma T_w^4}{{\rho }_{\text{abl}}{Q}^{\ast }}=\frac{3.01\times 10^6}{270\times 10^7}=1.11\times 10^{-3}\text{m/s}=1.11\text{mm/s}.$

WHY only $R_n$ matters here: in $\dot{q}\propto \sqrt{\rho /R_n}{v}^3$, doubling $R_n$ leaves $\rho$ and $v$ alone, so only the $1/\sqrt{R_n}$ factor changes. $\frac{{\dot{q}}_{\text{new}}}{{\dot{q}}_{\text{old}}}=\sqrt{\frac{R_{n,\text{old}}}{R_{n,\text{new}}}}=\sqrt{\frac{0.25}{0.50}}=\sqrt{0.5}=\mathrm{0.707.}$ Heating drops to ~71% — a ~29% reduction just by rounding the nose more. This is Blunt Body Aerodynamics in one line, and why capsules are round.

WHAT we compute: the temperature a purely radiating surface must reach to shed $10{\text{MW/m}}^2$. $T_w={\left(\frac{10^7}{0.85\times 5.67\times 10^{-8}}\right)}^{1/4}={\left(\frac{10^7}{4.82\times 10^{-8}}\right)}^{1/4}=(2.075\times 10^{14}{)}^{1/4}.$ $T_w\approx 3800\text{K}.$ DECISION: $3800\text{K}$ is far above the ~$1900\text{K}$ silica melting point. A re-radiating tile cannot survive — you must use an ablator, whose $\dot{m}{Q}^{\ast }$ blocking term removes energy no solid can radiate away. This is the quantitative heart of "reusable is not always better".

WHY multiply rate by time: if thickness burns at a steady $\dot{s}$, total loss is just $\dot{s}\times t$ (a rectangle of "rate × time"). $\Delta s=\dot{s}t=1.11\times 10^{-3}\times 90=0.0999\text{m}\approx 100\text{mm}.$ With a $50\%$ margin: $100\times 1.5=150\text{mm}$. So the heat shield must be at least 15 cm thick — real deep-space shields (e.g. Stardust-class PICA) are indeed this order of thickness.

WHY divide the scalings: since ${\dot{q}}_{\text{conv}}\propto v^3$ and ${\dot{q}}_{\text{rad}}\propto v^8$, their ratio scales as $v^8/v^3=v^5$. They start equal at $v_1$, so: $\frac{{\dot{q}}_{\text{rad}}}{{\dot{q}}_{\text{conv}}}{|}_{v_2}={\left(\frac{v_2}{v_1}\right)}^5={\left(\frac{11}{7}\right)}^5.$ ${\left(\frac{11}{7}\right)}^5=(1.5714{)}^5\approx \mathrm{9.57.}$ Radiative heating becomes nearly 10× larger than convective — this is exactly why deep-space return demands high-$Q^{\ast }$ ablators like PICA, not tiles, and why radiative blockage matters. See Mars Entry Descent and Landing.

WHY only $R_n$ enters: identical $\rho$, $v$ ⇒ ratio is set entirely by $1/\sqrt{R_n}$. $\frac{{\dot{q}}_{\text{LE}}}{{\dot{q}}_{\text{belly}}}=\sqrt{\frac{R_{n,\text{belly}}}{R_{n,\text{LE}}}}=\sqrt{\frac{1.0}{0.03}}=\sqrt{33.3}=\mathrm{5.77.}$ The leading edge sees ~5.8× the heat flux of the belly. Silica tiles (~$1900\text{K}$ limit) cannot take that; hence the sharp, hottest spots use RCC (survives $>1600{}^{\circ }\text{C}\approx 1900\text{K}$ and beyond). The Space Shuttle Columbia Accident was an RCC leading-edge breach — proof of how critical this hottest zone is.

One-line why for each level ::: L1 knows the families; L2 plugs into $\frac{1}{2}{v}^2$, Sutton–Graves, and $T_w=(\dot{q}/\epsilon \sigma {)}^{1/4}$; L3 balances ablation energy and the $1/\sqrt{R_n}$ blunt-body law; L4 shows radiation has a hard temperature ceiling that forces ablation; L5 handles $v^5$ crossover and per-zone material selection.