Ablative materials — phenolic-impregnated carbon ablator (PICA), AVCOAT, SLA: Worked examples

Formula The surface energy balance (re-stated from zero)

Heat arrives at the surface. It has only three places to go once it lands (any stored heat is bundled into the recession terms):

\overset{q}{˙}_{in}^{''} = carried off by escaping mass \overset{m}{˙}^{''} Q^{*} + radiated back to space ε σ T_{w}^{4} + conducted into the cold structure \overset{q}{˙}_{co n d}^{''}

Every symbol, in plain words:

$\overset{q}{˙}_{in}^{''}$ — the heat flux delivered to the surface by the shock layer, in watts per square metre ( $W/m^{2}$ ). It is the gross incoming number. It is NOT the same as $\overset{q}{˙}_{n e t}^{''}$ : part of $\overset{q}{˙}_{in}^{''}$ immediately leaves again as radiation and conduction, and only what remains drives ablation. Keep these two names strictly separate.
$\overset{q}{˙}_{n e t}^{''}$ — the net absorbed flux available to drive ablation, i.e. what is left after the surface has radiated and conducted: $\overset{q}{˙}_{n e t}^{''} \equiv \overset{q}{˙}_{in}^{''} - ε σ T_{w}^{4} - \overset{q}{˙}_{co n d}^{''} = \overset{m}{˙}^{''} Q^{*}$ . So $\overset{q}{˙}_{n e t}^{''}$ is exactly the term that shows up as $\overset{m}{˙}^{''} Q^{*}$ . When a problem hands you $\overset{q}{˙}_{n e t}^{''}$ directly, it has already done the radiation/conduction subtraction for you — do not subtract again.
$\overset{m}{˙}^{''}$ — kilograms of shield material leaving each square metre each second ( $kg/m^{2} s$ ). This is the recession rate in mass terms.
$Q^{*}$ — the effective heat of ablation: how many joules each kilogram of departing material carries away ( $J/kg$ ). Big $Q^{*}$ = a very "thirsty" material.
$ε$ — emissivity, a plain number between 0 and 1 saying how good a radiator the hot char is ( $ε = 1$ is a perfect radiator).
$σ$ — the Stefan–Boltzmann constant, $5.670 \times 1 0^{- 8} W/m^{2} K^{4}$ . It converts a temperature to radiated power.
$T_{w}$ — the wall (surface) temperature in kelvin ( $K$ ). Note the fourth power: double the temperature, radiate $2^{4} = 16$ times more.
$\overset{q}{˙}_{co n d}^{''}$ — the flux that gets conducted inward toward the crew/electronics, in $W/m^{2}$ . It carries the dot/double-prime for the same reason as every other flux. We want this near zero; " $\overset{q}{˙}_{co n d}^{''} \approx 0$ " in a problem means we're idealising conduction away, not that the symbol changed meaning.
$t$ — the elapsed time in seconds ( $s$ ) over which the heating acts.
$ρ$ — the density of the (charred) shield material, in kilograms per cubic metre ( $kg/m^{3}$ ): the mass packed into each cubic metre. A $1 m^{2}$ slab of thickness $s$ metres therefore holds mass $ρ s$ per unit area.
$s$ — the recession depth: how many metres of thickness the surface has been eaten back, $s = m^{''} / ρ$ (with $m^{''}$ the total mass lost per area).
$A$ — the shield forebody area in square metres ( $m^{2}$ ), used to turn a per-area loss into a total mass $M = m^{''} A$ (used in Cell G).
$M$ — the total ablated mass over the whole shield, in kilograms ( $kg$ ). No double-prime because it is already summed over the area.
$T_{p y r o}$ — the pyrolysis onset temperature: the wall temperature (roughly $600 K$ for phenolic) below which the resin will not decompose, so no char or gas — hence no ablation. See Pyrolysis & Char Yield.

The scenario matrix

Every worked example below is tagged with the cell it fills. If you can do all of these, the topic cannot ambush you.

Cell	The scenario class	What breaks / what to watch
A	Baseline: mass-loss from a net absorbed flux	plain plug-in, get $\overset{m}{˙}^{''}$ and depth $s$
B	Full balance: split $\overset{q}{˙}_{in}^{''}$ into radiated + ablated + conducted	none of the three terms is zero
C	Zero-ablation / radiative-equilibrium limit	$\overset{m}{˙}^{''} = 0$ : surface just glows, no material lost
D	Degenerate low flux: is $Q^{*}$ even reached?	flux too weak to pyrolyze — pick reusable TPS instead
E	Burn-through limit: shield too thin	recession depth $s$ exceeds thickness → failure
F	Material choice by density: same job, which is lighter?	compare $s = m^{''} / ρ$ across PICA / AVCOAT / SLA
G	Real word problem — Mars EDL (moderate flux, big area, long time)	integral heat load, area $A$ , total mass $M$
H	Exam twist — radiated fraction rises as $T_{w}$ climbs	non-linear $T^{4}$ ; find crossover temperature
I	Conduction-dominated regime: bad insulator / thin char	large $\overset{q}{˙}_{co n d}^{''}$ steals the balance → structure cooks

We use $σ = 5.670 \times 1 0^{- 8} W/m^{2} K^{4}$ throughout.

Cell A — Baseline mass loss and recession depth

Worked example A slab of PICA under a steady net absorbed flux

A PICA tile absorbs a net flux of $\overset{q}{˙}_{n e t}^{''} = 2 MW/m^{2}$ (already the radiation/conduction-subtracted value, so it goes straight into $\overset{m}{˙}^{''} Q^{*}$ ) for $t = 100 s$ . Its heat of ablation is $Q^{*} = 20 MJ/kg$ and char density $ρ = 270 kg/m^{3}$ . How much mass per area is lost, and how deep does the surface recede?

Forecast: guess the recession — a few millimetres? A few centimetres? Ten centimetres?

Total energy absorbed per area $= \overset{q}{˙}_{n e t}^{''} t = (2 \times 1 0^{6}) (100) = 2 \times 1 0^{8} J/m^{2}$ . Why this step? Power per area × time = energy per area. That is the total "heat bill" one square metre must handle.
Mass lost per area $m^{''} = \frac{q ˙ _{n e t}^{''} t}{Q ^{*}} = \frac{2 \times 1 0 ^{8}}{20 \times 1 0 ^{6}} = 10 kg/m^{2}$ . Why this step? Each kilogram carries away $Q^{*}$ joules, so total energy ÷ energy-per-kg = kilograms.
Recession depth $s = \frac{m ^{''}}{ρ} = \frac{10}{270} = 0.0370 m = 3.70 cm$ . Why this step? A column of material $s$ metres tall and $1 m^{2}$ wide has mass $ρ s$ ; set that equal to $m^{''}$ and solve for $s$ .

Verify: Units: $(kg/m^{2}) / (kg/m^{3}) = m$ ✓. Sanity: 3.7 cm of eaten shield after 100 s of megawatt heating is the textbook PICA figure — so we'd design thicker than 3.7 cm plus insulation margin.

Cell B — The full three-way split

Worked example Where does the incoming heat actually go?

A Dragon-class PICA shield sees $\overset{q}{˙}_{in}^{''} = 3.0 MW/m^{2}$ incoming. The char sits at $T_{w} = 2500 K$ with emissivity $ε = 0.9$ . Measured conduction into the structure is $\overset{q}{˙}_{co n d}^{''} = 0.20 MW/m^{2}$ . With $Q^{*} = 24 MJ/kg$ , find the mass-loss rate $\overset{m}{˙}^{''}$ and the percentage of incoming heat each mechanism handles.

Forecast: which term is biggest — radiation, ablation, or conduction?

Radiated flux $ε σ T_{w}^{4} = 0.9 \times (5.670 \times 1 0^{- 8}) \times (2500)^{4}$ . Compute $(2500)^{4} = 3.906 \times 1 0^{13}$ , so $ε σ T_{w}^{4} = 0.9 \times 5.670 \times 1 0^{- 8} \times 3.906 \times 1 0^{13} \approx 1.993 \times 1 0^{6} W/m^{2} \approx 1.99 MW/m^{2}$ . Why this step? Stefan–Boltzmann Radiation Law: a glowing surface exports power $\propto T^{4}$ . This is the "free" cooling we get just for being hot.
Solve the balance for the ablated flux (this is $\overset{q}{˙}_{n e t}^{''}$ ): $\overset{m}{˙}^{''} Q^{*} = \overset{q}{˙}_{n e t}^{''} = \overset{q}{˙}_{in}^{''} - ε σ T_{w}^{4} - \overset{q}{˙}_{co n d}^{''} = 3.00 - 1.99 - 0.20 = 0.81 MW/m^{2}$ . Why this step? Rearranging the master equation — whatever radiation and conduction don't take, the escaping mass must. Notice this is precisely the $\overset{q}{˙}_{n e t}^{''}$ that Cell A was handed pre-computed.
Mass-loss rate $\overset{m}{˙}^{''} = \frac{0.81 \times 1 0 ^{6}}{24 \times 1 0 ^{6}} = 0.0338 kg/m^{2} s = 33.8 g/m^{2} s$ . Why this step? Ablated flux ÷ energy-per-kg = kilograms leaving per second.
Shares of the incoming 3.0 MW/m²: radiation $1.99/3.0 = 66.4%$ , ablation $0.81/3.0 = 27.0%$ , conduction $0.20/3.0 = 6.7%$ . Why this step? Answers the exam favourite "what fraction is re-radiated?".

Verify: $66.4 + 27.0 + 6.7 = 100.1%$ (rounding) ✓. Radiation dominates at high $T_{w}$ — exactly why we love a hot, stable char and hate spallation that removes it.

Cell C — The radiative-equilibrium limit ( $\overset{m}{˙}^{''} = 0$ )

Worked example Radiative equilibrium temperature

Suppose $\overset{q}{˙}_{in}^{''} = 0.50 MW/m^{2}$ , $ε = 0.9$ , and (idealised) $\overset{q}{˙}_{co n d}^{''} \approx 0$ . If no ablation occurs, what surface temperature does the shield settle at?

Forecast: hotter or cooler than the 2500 K of Cell B?

Set the balance with $\overset{m}{˙}^{''} = 0, \overset{q}{˙}_{co n d}^{''} = 0$ : $\overset{q}{˙}_{in}^{''} = ε σ T_{w}^{4}$ . Why this step? If nothing is ablated or conducted, all incoming heat must leave by radiation — that is the definition of radiative equilibrium.
Solve for $T_{w}$ : $T_{w} = (\frac{q ˙ _{in}^{''}}{ε σ})^{1/4} = (\frac{0.50 \times 1 0 ^{6}}{0.9 \times 5.670 \times 1 0 ^{- 8}})^{1/4}$ . Why the fourth root? Radiation scales as $T^{4}$ ; to invert a fourth power you take a fourth root — the only operation that undoes it.
Number: inside the bracket $= \frac{0.50 \times 1 0 ^{6}}{5.103 \times 1 0 ^{- 8}} = 9.798 \times 1 0^{12}$ ; its fourth root $\approx 1770 K$ . Why this step? $T_{w} = (9.798 \times 1 0^{12})^{1/4} \approx 1770 K$ .

Verify: Plug back: $0.9 \times 5.670 \times 1 0^{- 8} \times (1770)^{4} \approx 0.50 \times 1 0^{6}$ ✓. It is cooler than 2500 K — a lower flux needs a lower glowing temperature, no material sacrificed. Below this regime, a reusable TPS beats an ablator.

The picture below plots the black radiation curve $ε σ T_{w}^{4}$ climbing with temperature and the flat red line of incoming flux; where they cross is the equilibrium temperature we just solved for. Read it as: at any $T_{w}$ left of the red dot the surface can't radiate fast enough (net heating, so it warms up); right of it it over-radiates (net cooling, so it settles back). The crossing is the only stable resting point.

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

Cell D — Degenerate low-flux: is ablation even triggered?

Worked example Too gentle to char

A satellite in a low-energy re-entry sees only $\overset{q}{˙}_{in}^{''} = 0.08 MW/m^{2}$ . Phenolic pyrolysis needs the wall above the onset temperature $T_{p y r o} \approx 600 K$ to begin producing char and gas. With $ε = 0.9$ , does ablation switch on?

Forecast: does the char ever form, or is this a job for reusable tiles?

Find the radiative-equilibrium temperature (as in Cell C): $T_{w} = (\frac{0.08 \times 1 0 ^{6}}{0.9 \times 5.670 \times 1 0 ^{- 8}})^{1/4}$ . Why this step? If the surface can radiate away all the heat while staying below $T_{p y r o}$ , no material is lost.
Number: bracket $= \frac{8 \times 1 0 ^{4}}{5.103 \times 1 0 ^{- 8}} = 1.568 \times 1 0^{12}$ ; fourth root $\approx 1119 K$ .
Compare to threshold: $1119 K > T_{p y r o} = 600 K$ . Why this step? The equilibrium glow temperature exceeds the pyrolysis onset, so the surface does get hot enough to char — ablation still triggers, just weakly.

Verify: Even a "gentle" 80 kW/m² drives the surface past red heat. The takeaway: ablation is remarkably easy to trigger; the design question is rate and total mass lost, not whether it happens. For truly mild, repeatable heating (Shuttle LEO return) you still prefer reusable tiles — because you don't want to lose mass every flight, not because ablation wouldn't start. See the "reusable vs ablative" trade in the parent.

Cell E — Burn-through: when the shield is too thin

Worked example Does 3 cm survive Cell A's load?

Reuse Cell A's PICA: it loses $m^{''} = 10 kg/m^{2}$ , giving recession $s = 3.70 cm$ . Engineering also demands an insulation margin of $2.0 cm$ of unpyrolyzed material behind the char to keep $\overset{q}{˙}_{co n d}^{''}$ acceptable. Is a 5.0 cm tile safe? Is a 3.0 cm tile safe?

Forecast: which thickness burns through?

Required thickness $= s + margin = 3.70 + 2.0 = 5.70 cm$ . Why this step? The char eats $3.70 cm$ ; behind it you still need virgin material as an insulator, or the structure cooks.
5.0 cm tile: $5.0 < 5.70$ → fails the margin (survives recession but leaves only $1.30 cm$ of insulator, under the $2.0 cm$ requirement). Why this step? Surviving recession is necessary but not sufficient; the margin is the real design constraint.
3.0 cm tile: $3.0 < 3.70$ → catastrophic burn-through — the recession front reaches the structure with material to spare. Why this step? If $s >$ thickness, the ablation front punches through entirely. This is the failure mode.

Verify: Order of severity: 3.0 cm (burn-through) is worse than 5.0 cm (thin margin), which is worse than 5.70 cm (safe) ✓. This is why real shields carry generous margins — recession estimates have uncertainty and spallation can accelerate loss. Conclusion: neither the 3.0 cm nor the 5.0 cm tile is acceptable; you must build at least 5.70 cm.

Cell F — Same job, which material is lighter?

Worked example Comparing PICA, AVCOAT, SLA for one heat load

A mission absorbs $\overset{q}{˙}_{n e t}^{''} = 1.5 MW/m^{2}$ for $t = 80 s$ . Assume (for a clean comparison) all three materials share $Q^{*} = 18 MJ/kg$ but differ in density: PICA $270$ , SLA $260$ , AVCOAT $500 kg/m^{3}$ . Compare recession depth $s$ and lost mass per area.

Forecast: which material recedes deepest? Which loses most mass?

Mass lost per area is the same for all three: $m^{''} = \frac{q ˙ _{n e t}^{''} t}{Q ^{*}} = \frac{( 1.5 \times 1 0 ^{6} ) ( 80 )}{18 \times 1 0 ^{6}} = 6.667 kg/m^{2}$ . Why this step? $m^{''}$ depends only on energy and $Q^{*}$ , not on density. Same $Q^{*}$ → same kilograms sacrificed.
Recession depths $s = m^{''} / ρ$ :
- PICA: $6.667/270 = 0.0247 m = 2.47 cm$
- SLA: $6.667/260 = 0.0256 m = 2.56 cm$
- AVCOAT: $6.667/500 = 0.0133 m = 1.33 cm$ Why this step? Lighter material spreads the same lost mass over a taller column, so it recedes deeper. Denser AVCOAT recedes least.
Interpret: AVCOAT recedes least but you paid for it with density — for the same $s$ -margin it weighs almost twice as much per cm. Why this step? The real figure of merit is total shield mass, and every kilogram costs launch energy.

Verify: $s_{PICA} < s_{SLA}$ because $270 > 260$ ✓; AVCOAT deepest-surviving but heaviest ✓. This is exactly the parent's mistake-buster: low density + high $Q^{*}$ wins, which is why PICA and SLA dominate mass-critical missions.

Cell G — Real word problem: Mars EDL mass budget

Worked example Sizing an SLA heat shield for Mars entry

A Mars aeroshell has a forebody area $A = 15 m^{2}$ . It experiences a moderate average net absorbed flux $\overset{q}{˙}_{n e t}^{''} = 0.35 MW/m^{2}$ but for a long duration $t = 90 s$ (long because the thin CO₂ atmosphere decelerates gently). Using SLA with $Q^{*} = 15 MJ/kg$ and $ρ = 260 kg/m^{3}$ , find (a) recession depth $s$ and (b) total ablated mass $M$ over the whole forebody.

Forecast: will Mars eat more or fewer centimetres than the megawatt cases above?

Mass lost per area $m^{''} = \frac{q ˙ _{n e t}^{''} t}{Q ^{*}} = \frac{( 0.35 \times 1 0 ^{6} ) ( 90 )}{15 \times 1 0 ^{6}} = 2.10 kg/m^{2}$ . Why this step? Same master formula — Mars just supplies a smaller flux over a longer time.
Recession depth $s = \frac{m ^{''}}{ρ} = \frac{2.10}{260} = 0.00808 m = 0.81 cm$ . Why this step? Only ~8 mm lost — Mars heating is mild, so a thin ultralight SLA layer suffices. This is the whole reason SLA is chosen for Mars.
Total ablated mass $M = m^{''} \times A = 2.10 \times 15 = 31.5 kg$ . Why this step? Per-area loss times the shield area $A$ gives the total mass sacrificed — a line item in the mass budget. This is the one case where the area $A$ enters.

Verify: Units for $s$ : $(kg/m^{2}) / (kg/m^{3}) = m$ ✓. Units for $M$ : $(kg/m^{2}) (m^{2}) = kg$ ✓. Compare to Cell A's 3.70 cm — Mars recedes far less (0.81 cm), confirming the intuition that Mars entry is flux-moderate, area- and duration-driven, so ultralight SLA is the mass-efficient pick.

Cell H — Exam twist: the $T^{4}$ crossover

Worked example At what wall temperature does radiation carry half the incoming flux?

Incoming $\overset{q}{˙}_{in}^{''} = 2.0 MW/m^{2}$ , emissivity $ε = 0.9$ . Find the wall temperature $T_{w}$ at which the radiated flux equals half the incoming flux (i.e. $1.0 MW/m^{2}$ ).

Forecast: roughly 2000 K? 2200 K? Higher?

Set the condition: $ε σ T_{w}^{4} = 0.5 \overset{q}{˙}_{in}^{''} = 1.0 \times 1 0^{6} W/m^{2}$ . Why this step? We're asking "which temperature radiates exactly this much?" — invert the Stefan–Boltzmann law.
Solve: $T_{w} = (\frac{1.0 \times 1 0 ^{6}}{0.9 \times 5.670 \times 1 0 ^{- 8}})^{1/4}$ . Why the fourth root again? Same reason as Cell C — it's the only operation that undoes a fourth power.
Number: bracket $= \frac{1.0 \times 1 0 ^{6}}{5.103 \times 1 0 ^{- 8}} = 1.960 \times 1 0^{13}$ ; fourth root $\approx 2103 K$ . Why this step? $T_{w} = (1.960 \times 1 0^{13})^{1/4} \approx 2103 K$ .

Verify: Plug back: $0.9 \times 5.670 \times 1 0^{- 8} \times (2103)^{4} \approx 1.0 \times 1 0^{6}$ ✓. Above 2103 K, radiation carries more than half — the char's glow does most of the work, and the ablation rate falls. This is the physical reason a stable, high-emissivity char is worth protecting.

Cell I — The conduction-dominated regime (the edge case that cooks the crew)

Worked example A degraded char lets conduction dominate

The same shield of Cell B sees $\overset{q}{˙}_{in}^{''} = 3.0 MW/m^{2}$ at $T_{w} = 2500 K$ , $ε = 0.9$ (so $ε σ T_{w}^{4} \approx 1.99 MW/m^{2}$ ), $Q^{*} = 24 MJ/kg$ . But now the char has partially spalled and conduction has jumped to $\overset{q}{˙}_{co n d}^{''} = 0.90 MW/m^{2}$ . Find $\overset{m}{˙}^{''}$ , its share of the incoming heat, and compare the conduction share to Cell B.

Forecast: does ablation speed up or slow down when conduction hogs the balance?

Ablated flux $\overset{q}{˙}_{n e t}^{''} = \overset{q}{˙}_{in}^{''} - ε σ T_{w}^{4} - \overset{q}{˙}_{co n d}^{''} = 3.00 - 1.99 - 0.90 = 0.11 MW/m^{2}$ . Why this step? The master balance is a fixed pie: radiation and conduction take their slices first, and ablation gets only the crumbs left over.
Mass-loss rate $\overset{m}{˙}^{''} = \frac{0.11 \times 1 0 ^{6}}{24 \times 1 0 ^{6}} = 0.00458 kg/m^{2} s = 4.58 g/m^{2} s$ . Why this step? Ablated flux ÷ energy-per-kg. Notice it is roughly 7× smaller than Cell B's $33.8 g/m^{2} s$ .
Conduction share $= 0.90/3.0 = 30.0%$ , versus only $6.7%$ in Cell B. Why this step? Quantifies the danger: nearly a third of the incoming heat is now marching toward the structure instead of $\frac{1}{15}$ .

Verify: Slices sum: $1.99 + 0.11 + 0.90 = 3.00 MW/m^{2} = \overset{q}{˙}_{in}^{''}$ ✓. Counter-intuitively, ablation slows when conduction dominates — because ablation only feeds on what radiation and conduction leave behind. That is exactly why spallation is bad: losing the char both cuts the protective mass and lets conduction surge. Keep the char, keep $\overset{q}{˙}_{co n d}^{''}$ tiny.

Recap: the matrix, closed

Recall Which cell tests what?

Cell A tests plain recession ::: mass = $\overset{q}{˙}_{n e t}^{''} t / Q^{*}$ , depth = $m^{''} / ρ$ . Cell B tests the full three-way split ::: radiation + ablation + conduction = $\overset{q}{˙}_{in}^{''}$ . Cell C tests the zero-ablation limit ::: $\overset{q}{˙}_{in}^{''} = ε σ T_{w}^{4}$ , solve $T_{w}$ by fourth root. Cell D tests the degenerate low flux ::: check equilibrium $T_{w}$ against $T_{p y r o}$ . Cell E tests burn-through ::: fail if recession + margin exceeds thickness. Cell F tests density choice ::: same $m^{''}$ , deeper $s$ for lighter material. Cell G tests a real Mars budget ::: total mass $M = m^{''} \times A$ . Cell H tests the $T^{4}$ crossover ::: find $T_{w}$ where radiation equals a target flux. Cell I tests the conduction-dominated regime ::: big $\overset{q}{˙}_{co n d}^{''}$ starves ablation and cooks the structure.