Exercises — Ablative materials — phenolic-impregnated carbon ablator (PICA), AVCOAT, SLA

AVCOAT, at about ==$500{\text{kg/m}}^3$==. PICA (~270) and SLA (~260) are both far lighter — remember the mnemonic "Mars and comets travel light; the Moon ride is heavier." AVCOAT carried Apollo and Orion crews home from the Moon on a big curved shield, where mechanical robustness (the honeycomb) mattered more than shaving every last kilogram.

1. Endothermic pyrolysis — the phenolic resin decomposes, and breaking chemical bonds absorbs heat.
2. Pyrolysis-gas blowing (transpiration cooling) — the gases flow outward, thicken the hot boundary layer, and block heat from conducting inward.
3. Char re-radiation — the porous carbon char glows and radiates energy back to space, proportional to $T^4$ (see Stefan–Boltzmann Radiation Law).

False. The dominant processes are endothermic — they absorb energy. Combustion (oxidation) would release energy and even eat the protective char, which is undesirable. Ablation is mostly cooling, not burning.

$m^{\prime \prime }=\frac{{\dot{q}}_{net}t}{Q^{\ast }}=\frac{(2\times 10^6)(120)}{20\times 10^6}=\frac{2.4\times 10^8}{2\times 10^7}=12{\text{kg/m}}^2.$ Why: total energy absorbed over the area (${\dot{q}}_{net}t$) divided by the energy each kilogram can handle ($Q^{\ast }$) gives the kilograms sacrificed.

$s=\frac{m^{\prime \prime }}{\rho }=\frac{12}{270}\approx 0.0444\text{m}=4.44\text{cm}.$ Why: kilograms lost per square metre, divided by kilograms per cubic metre, leaves metres — a thickness. You would design the shield thicker than this plus an insulating margin.

${\dot{q}}_{rad}=0.9\times 5.67\times 10^{-8}\times (3000{)}^4.$ First $(3000{)}^4=8.1\times 10^{13}$. Then ${\dot{q}}_{rad}=0.9\times 5.67\times 10^{-8}\times 8.1\times 10^{13}\approx 4.13\times 10^6{\text{W/m}}^2\approx 4.1{\text{MW/m}}^2.$ Why this tool: radiation is the only way the surface dumps heat straight back to empty space without any material contact, and it grows as the fourth power of temperature — so a hot char is a spectacularly effective radiator. That single $T^4$ term is why letting the char get hot is a feature, not a bug.

Total absorbed energy per area is the same for both: $E={\dot{q}}_{net}t=1.5\times 10^6\times 200=3\times 10^8{\text{J/m}}^2$.

Mass lost $m^{\prime \prime }=E/Q^{\ast }$: $m_A^{\prime \prime }=\frac{3\times 10^8}{15\times 10^6}=20{\text{kg/m}}^2,m_B^{\prime \prime }=\frac{3\times 10^8}{30\times 10^6}=10{\text{kg/m}}^2.$ Material B loses less mass (its higher $Q^{\ast }$ wins).

Recession $s=m^{\prime \prime }/\rho$: $s_A=\frac{20}{260}\approx 0.0769\text{m}=7.69\text{cm},s_B=\frac{10}{500}=0.020\text{m}=2.0\text{cm}.$ Material B also recedes less. Here they agree — B is better on both counts.

Why they needn't agree: recession depends on $Q^{\ast }\rho$ in the denominator ($s=E/(Q^{\ast }\rho )$), while mass loss depends on $Q^{\ast }$ alone. A very light material with modest $Q^{\ast }$ could lose more mass yet recede a similar depth. The two metrics answer different questions: mass loss = "how much launch weight did I spend?", recession = "how thick must the tile be?"

The pyrolysis gases push outward against the incoming hot flow, fattening the insulating cushion of gas (the boundary layer) between the shock-heated air and the wall. A thicker cushion means a gentler temperature gradient at the wall, and conduction is driven by that gradient. So more of the incoming heat is turned away (or re-radiated) and less conducts inward — the ${\dot{q}}_{cond}$ term in the wall energy balance shrinks. This is transpiration cooling, and it is why we want the resin to give off gas as it pyrolyses, not just leave char.

(a) Absorbed energy per area $E={\dot{q}}_{net}t=0.8\times 10^6\times 400=3.2\times 10^8{\text{J/m}}^2$. $s=\frac{E}{Q^{\ast }\rho }=\frac{3.2\times 10^8}{(12\times 10^6)(260)}=\frac{3.2\times 10^8}{3.12\times 10^9}\approx 0.1026\text{m}\approx 10.3\text{cm}.$

(b) Mass per area $m^{\prime \prime }=E/Q^{\ast }=3.2\times 10^8/12\times 10^6\approx 26.67{\text{kg/m}}^2$. Over $15{\text{m}}^2$: $M=m^{\prime \prime }\times A=26.67\times 15=400\text{kg}.$

(c) SLA is ultralight and flexible, ideal for large-area, moderate-flux, long-duration ${\text{CO}}_2$ entries where mass efficiency over a big aeroshell dominates; PICA's rigid tiles shine at high flux and would be over-engineered (and gap-filling a whole Mars aeroshell in tiles is impractical).

(i) AVCOAT — high integral heat load over a big curved shield; the gunned honeycomb gives robustness and a glassy shear-resistant melt. (ii) SLA-561V — moderate flux, large area, mass-critical, ${\text{CO}}_2$ atmosphere. (iii) PICA — extreme flux at (relatively) low pressure; its carbon skeleton keeps shape and $Q^{\ast }$ is very high. (Stardust used PICA at the fastest-ever Earth re-entry.)

(a) Re-radiation (Stefan–Boltzmann): $(3200{)}^4=1.048576\times 10^{14}$. ${\dot{q}}_{rad}=0.85\times 5.67\times 10^{-8}\times 1.048576\times 10^{14}\approx 5.053\times 10^6{\text{W/m}}^2\approx 5.05{\text{MW/m}}^2.$

(b) Net absorbed = incoming − re-radiated − conducted: ${\dot{q}}_{net}=6.0-5.05-0.30\approx 0.647{\text{MW/m}}^2=6.47\times 10^5{\text{W/m}}^2.$ Notice how much of the 6 MW/m² is simply radiated straight back — the hot char is doing most of the defending.

(c) Mass-loss rate from recession: ${\dot{m}}^{\prime \prime }=\rho \dot{s}=270\times (0.20\times 10^{-3})=0.054{\text{kg/m}}^2\text{s}$.

(d) Effective heat of ablation: $Q^{\ast }=\frac{{\dot{q}}_{net}}{{\dot{m}}^{\prime \prime }}=\frac{6.47\times 10^5}{0.054}\approx 1.198\times 10^7\text{J/kg}\approx 12.0\text{MJ/kg}.$

Why this ties everything together: we used the Stefan–Boltzmann Radiation Law for re-radiation, the wall energy balance for the net flux, the definition of density to turn recession into mass loss, and finally the definition of $Q^{\ast }$. Every quantity earns its place in the same equation.

Two of the three cooling tricks depend on a stable char surviving after pyrolysis:

• Re-radiation ($\epsilon \sigma T^4$) needs a porous refractory carbon layer to glow and dump heat to space. No char, no radiator.
• Insulation of the virgin layer needs that same char skeleton to sit in place and slow conduction.

At 10% char yield, ~90% of the resin boils away as gas, leaving almost no char. You would still get some transpiration-gas blowing and endothermic bond-breaking, but you would lose the dominant radiative defence (recall in L5.1 that radiation carried ~5 of 6 MW/m²). The surface would recede fast, exposing fresh virgin material continuously, and shield thickness would balloon. High char yield is precisely why cross-linked aromatic phenolic resin is the workhorse — see also Carbon-Carbon Composites & RCC for the pure-carbon extreme.

Can you, without notes, ...

• State which of $m^{\prime \prime }$ and $s$ depends on $\rho$? ::: $s=m^{\prime \prime }/\rho$ depends on density; $m^{\prime \prime }={\dot{q}}_{net}t/Q^{\ast }$ does not.
• Say which flux term dominates the wall defence at ~3000 K? ::: Re-radiation $\epsilon \sigma T_w^4$ — it scales as $T^4$.
• Match PICA / AVCOAT / SLA to their regimes? ::: PICA = high flux/low pressure; AVCOAT = big curved shields/high integral load; SLA = low flux/large area/Mars.