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

Before any numbers, let us re-state the three tools in plain words so every symbol is earned.

Formula The three tools (restated from zero)

Tool 1 — Convective heating (Sutton–Graves). $\overset{q}{˙}_{conv} = k \frac{ρ}{R _{n}} v^{3}$

$\overset{q}{˙}_{conv}$ = heat power arriving on one square metre of the nose, in watts per square metre ( $W/m^{2}$ ). Think "how fast heat pours onto a stamp-sized patch."
$ρ$ = air density where the vehicle is, in kg/m³ (thicker air → more molecules to bang into → more heat).
$R_{n}$ = nose radius in metres: the radius of the ball the nose looks like. Bigger ball = gentler.
$v$ = speed relative to the air, in m/s.
$k \approx 1.7 \times 1 0^{- 4}$ — an empirical number that makes the formula match wind-tunnel data. Why a square-root on $ρ / R_{n}$ and a cube on $v$ ? Speed enters three times (energy per mass $\sim v^{2}$ , and the rate of mass hitting the wall $\sim v$ ), giving $v^{3}$ . The on $ρ / R_{n}$ is the boundary-layer scaling the parent derived.

Definition The hidden units of

$k$ (so the answer really lands in W/m²) The bracketed pieces carry units: $ρ / R_{n}$ has units $(kg m^{- 3}) / m = kg^{1/2} m^{- 2}$ , and $v^{3}$ has units $m^{3} s^{- 3}$ . Multiplying those gives $kg^{1/2} m s^{- 3}$ . To turn that into $W/m^{2} = kg s^{- 3}$ , the constant $k$ must itself carry units $[k] = \frac{kg s ^{- 3}}{kg ^{1/2} m s ^{- 3}} = kg^{1/2} m^{- 1} .$ So "SI units" for $k$ means $k = 1.7 \times 1 0^{- 4} kg^{1/2} m^{- 1}$ . This is why $k$ is an empirical constant: it silently absorbs the leftover half-power of mass and inverse metre that the physics scaling could not fix on its own.

Tool 2 — Re-radiation (Stefan–Boltzmann). A hot surface throws heat back out: $\overset{q}{˙}_{rad,out} = ε σ T_{w}^{4}$

$ε$ = emissivity, a pure number from $0$ (shiny, radiates poorly) to $1$ (perfect black radiator).
$σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ = Stefan–Boltzmann constant.
$T_{w}$ = wall temperature in kelvin (K). See Radiative Heat Transfer and Stefan–Boltzmann Law.

Tool 3 — Ablator energy balance. What arrives, minus what re-radiates, goes into eating material: $\overset{s}{˙} = \frac{q ˙ _{conv} - ε σ T _{w}^{4}}{ρ _{abl} Q ^{*}}$

$\overset{s}{˙}$ = recession rate: how fast the surface eats inward, in m/s.
$ρ_{abl}$ = density of the ablator (kg/m³).
$Q^{*}$ = effective heat of ablation (J/kg) — energy to remove one kilogram of shield.

Intuition When Tool 1 itself breaks (radiation-dominated entry)

Sutton–Graves (Tool 1) is a convective correlation: it assumes heat reaches the wall by hot gas touching it. At extreme speeds (roughly $v > 10 km/s$ , Mach numbers past ~25 — lunar and Mars return) the shock gas gets so hot it glows, and a second channel opens: the gas radiates heat straight at the wall. That radiative input $\overset{q}{˙}_{rad,in}$ is not in Tool 1 at all, and it scales roughly as $ρ^{a} R_{n}^{b} v^{c}$ with a much steeper power of $v$ (often $c \approx 8$ – $12$ , far above the convective $v^{3}$ ). So at very high speed, convective correlations under-predict the real load and you must add the radiative term — this is exactly the regime Ex 4 sits at the edge of, and why PICA (a high- $Q^{*}$ ablator) is mandatory there. See Radiative Heat Transfer and Stefan–Boltzmann Law and Reentry Aerothermodynamics.

The scenario matrix

Every problem on this page is one cell of the table below. If a cell is not covered, you would one day meet it unprepared — so we cover them all.

#	Cell class	What is extreme / tricky	Example
C1	Baseline convective heating	ordinary LEO capsule	Ex 1
C2	Small $R_{n}$ limit (sharp nose)	$R_{n} \to$ small → $\overset{q}{˙} \to$ huge	Ex 2
C3	Large $R_{n}$ / density-zero limit	$R_{n} \to \infty$ or $ρ \to 0$ → $\overset{q}{˙} \to 0$	Ex 3
C4	Speed scaling ( $v^{3}$ sensitivity, radiation onset)	double $v$ → ×8 heat; $v > 10$ km/s breaks Tool 1	Ex 4
C5	Radiative equilibrium wall temp	tile survival, invert $T^{4}$	Ex 5
C6	Ablator recession (all terms live)	subtract re-radiation, get $\overset{s}{˙}$	Ex 6
C7	Degenerate ablation ( $\overset{s}{˙} \to 0$ )	re-radiation ≥ incoming → no recession	Ex 7
C8	Real-world word problem (mass budget)	total shield mass for a Mars entry	Ex 8
C9	Exam twist (which TPS + why)	$T_{w}$ exceeds melting → forces ablator	Ex 9

Constants used throughout (memorise these): $k = 1.7 \times 1 0^{- 4} kg^{1/2} m^{- 1}, σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4} .$

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

Worked examples

Worked example Ex 1 — C1 Baseline: a LEO capsule

A capsule reenters at $v = 7.5 km/s$ where air density is $ρ = 1.0 \times 1 0^{- 3} kg/m^{3}$ . Nose radius $R_{n} = 1.0 m$ , $k = 1.7 \times 1 0^{- 4}$ . Find $\overset{q}{˙}_{conv}$ .

Forecast: Guess the order of magnitude first — millions of watts per square metre? Tens of thousands? Write down your guess before reading on.

Convert speed to SI: $v = 7500 m/s$ . Why this step? The formula is in SI; km/s would silently multiply the answer by $100 0^{3}$ .
Compute the density/radius factor: $ρ / R_{n} = 1 0^{- 3} /1.0 = 1 0^{- 3} = 0.03162$ . Why this step? This is the "how many molecules, how spread out" part — isolate it so the cube of $v$ is separate.
Cube the speed: $v^{3} = (7500)^{3} = 4.219 \times 1 0^{11}$ . Why this step? Speed is the dominant lever; getting it right matters most.
Multiply: $\overset{q}{˙}_{conv} = 1.7 \times 1 0^{- 4} \times 0.03162 \times 4.219 \times 1 0^{11} = 2.27 \times 1 0^{6} W/m^{2}$ .

Answer: $\overset{q}{˙}_{conv} \approx 2.27 MW/m^{2}$ .

Verify: Units: $[k] \cdot kg m^{- 3} / m \cdot (m/s)^{3} = kg^{1/2} m^{- 1} \cdot kg^{1/2} m^{- 2} \cdot m^{3} s^{- 3} = kg s^{- 3} = W/m^{2}$ . ✓ Sanity: a few $MW/m^{2}$ is the textbook figure for LEO return — see Reentry Aerothermodynamics. Your forecast of "millions" was right.

Worked example Ex 2 — C2 Sharp-nose limit: why capsules are round

Take Ex 1 but shrink the nose to $R_{n} = 0.01 m$ (a 1 cm sharp edge, like the Shuttle wing leading edge). Same $ρ$ , same $v$ . Find the new $\overset{q}{˙}$ and the ratio to Ex 1.

Forecast: $R_{n}$ dropped by ×100. Since $\overset{q}{˙} \propto 1/ R_{n}$ , guess the heating multiplier before computing.

New factor: $ρ / R_{n} = 1 0^{- 3} /0.01 = 0.1 = 0.3162$ . Why this step? Only $R_{n}$ changed; everything else in Ex 1 is reusable.
Ratio to Ex 1: $1.0/0.01 = 100 = 10$ . Why this step? $\overset{q}{˙} \propto 1/ R_{n}$ , so a ×100 smaller radius gives $100 = 10$ × the heat — the limit worth seeing.
New heating: $\overset{q}{˙} = 10 \times 2.27 MW/m^{2} = 2.27 \times 1 0^{7} W/m^{2} = 22.7 MW/m^{2}$ .

Answer: $22.7 MW/m^{2}$ — ten times the blunt case.

Verify: Push $R_{n} \to 0$ (truly sharp): $1/ R_{n} \to \infty$ , so $\overset{q}{˙} \to \infty$ . That is the mathematical proof that a razor nose roasts — exactly why leading edges need RCC, not silica tiles.

Worked example Ex 3 — C3 Vanishing limits: high altitude & big shield

Two degenerate checks. (a) The vehicle is still in near-vacuum, $ρ = 1.0 \times 1 0^{- 8} kg/m^{3}$ ( $R_{n} = 1$ m, $v = 7500$ m/s). (b) A giant inflatable shield, $R_{n} = 100 m$ (back to $ρ = 1 0^{- 3}$ ).

Forecast: Both should make heating small. Which drops faster for a factor-of- $1 0^{5}$ change — density or radius?

(a) Density factor: $1 0^{- 8} /1 = 1 0^{- 4}$ . Then $\overset{q}{˙} = 1.7 \times 1 0^{- 4} \times 1 0^{- 4} \times 4.219 \times 1 0^{11} = 7.17 \times 1 0^{3} W/m^{2}$ . Why this step? Shows the top-of-atmosphere limit: almost no molecules, almost no heat. As $ρ \to 0$ , $\overset{q}{˙} \to 0$ smoothly.
(b) Radius factor: $1 0^{- 3} /100 = 1 0^{- 5} = 3.162 \times 1 0^{- 3}$ . Then $\overset{q}{˙} = 1.7 \times 1 0^{- 4} \times 3.162 \times 1 0^{- 3} \times 4.219 \times 1 0^{11} = 2.27 \times 1 0^{5} W/m^{2}$ . Why this step? As $R_{n} \to \infty$ , $\overset{q}{˙} \to 0$ — the mathematical reason "blunter is cooler."

Answer: (a) $\approx 7.2 kW/m^{2}$ ; (b) $\approx 0.227 MW/m^{2}$ .

Verify: Density enters linearly under the root ( $ρ$ ); a $1 0^{- 5}$ drop in $ρ$ gives $1 0^{- 5} \approx 3 \times 1 0^{- 3}$ — same factor as a $1 0^{- 5}$ drop would give in $1/ R_{n}$ . So symmetry holds: both channels shut off as their variable heads to the gentle extreme. Case (a) is $1000 \times$ smaller than case (b) because $ρ$ fell $1 0^{5}$ while $R_{n}$ only rose $1 0^{2}$ .

Worked example Ex 4 — C4 Speed scaling: the

$v^{3}$ lever (and where Tool 1 breaks) A lunar-return capsule hits $v = 11 km/s$ instead of the LEO $7.5 km/s$ (same $ρ = 1 0^{- 3}$ , $R_{n} = 1$ ). By what factor does convective heating jump, and what is its value? Then note the caveat.

Forecast: Speed rose by $11/7.5 \approx 1.47$ . Because of $v^{3}$ , guess the heat multiplier.

Speed ratio: $11/7.5 = 1.4667$ . Why this step? Isolate the only thing that changed.
Cube it: $1.466 7^{3} = 3.155$ . Why this step? $\overset{q}{˙} \propto v^{3}$ , so the heating scales as the cube of the speed ratio — the single most important sensitivity in reentry.
New convective heating: $\overset{q}{˙}_{conv} = 3.155 \times 2.27 MW/m^{2} = 7.16 MW/m^{2}$ .

Answer: $\approx 3.16 \times$ higher, $\approx 7.16 MW/m^{2}$ — convective only.

Verify: A mere 47% faster tripled the heat. Caveat (the radiation edge case): at $11 km/s$ we are past $10 km/s$ , so the glowing-gas radiative input $\overset{q}{˙}_{rad,in}$ that Tool 1 omits now matters and can rival or exceed the convective value. The $7.16 MW/m^{2}$ is therefore a lower bound; the true load is larger, which is precisely why deep-space returns (see Specific Impulse and Energy Budgets) demand a high- $Q^{*}$ PICA ablator. Direct check: $(11000)^{3} / (7500)^{3} = 3.155$ . ✓

Worked example Ex 5 — C5 Radiative equilibrium: will a silica tile survive?

A reusable silica tile faces $\overset{q}{˙}_{conv} = 3.0 \times 1 0^{5} W/m^{2}$ on the Shuttle belly. Its coating has emissivity $ε = 0.85$ . If all incoming heat is re-radiated, find the steady wall temperature $T_{w}$ . (Tiles fail above ~ $1500 K$ .)

Forecast: Balance says $\overset{q}{˙}_{conv} = ε σ T_{w}^{4}$ . We invert a fourth power — will $T_{w}$ land under or over 1500 K?

Write the balance: incoming = outgoing radiation, $\overset{q}{˙}_{conv} = ε σ T_{w}^{4}$ . Why this step? "Steady" means the tile stops heating up when what it radiates equals what arrives — energy in = energy out.
Solve for $T_{w}$ : $T_{w} = (\frac{q ˙ _{conv}}{ε σ})^{1/4}$ . Why the fourth root? Radiation grows as $T^{4}$ ; to undo it we take the fourth root — the inverse operation, just as undoes a square.
Plug in: $\frac{3.0 \times 1 0 ^{5}}{0.85 \times 5.67 \times 1 0 ^{- 8}} = 6.226 \times 1 0^{12}$ . Fourth root: $T_{w} = (6.226 \times 1 0^{12})^{1/4} = 1580 K$ . Why this step? Substitute the numbers into the inverted formula and actually take the fourth root — this converts the abstract balance into a concrete kelvin value we can compare against the 1500 K survival limit.

Answer: $T_{w} \approx 1580 K$ .

Verify: Slightly above the 1500 K limit — this belly location is near the tile's edge of survival, which is exactly the real design margin story. Sanity on the fourth root: raising $ε$ from 0.85 toward 1 would lower $T_{w}$ (a better radiator runs cooler), confirming the black high- $ε$ coating choice.

Worked example Ex 6 — C6 Full ablator: recession with re-radiation live

A PICA shield ( $ρ_{abl} = 270 kg/m^{3}$ , $Q^{*} = 1.2 \times 1 0^{7} J/kg$ , $ε = 0.9$ ) sits at char temperature $T_{w} = 3000 K$ under $\overset{q}{˙}_{conv} = 8.0 \times 1 0^{6} W/m^{2}$ . Find the recession rate $\overset{s}{˙}$ .

Forecast: Some heat re-radiates first; the rest eats material. Guess whether re-radiation is a big or small slice of 8 MW/m².

Re-radiated flux: $ε σ T_{w}^{4} = 0.9 \times 5.67 \times 1 0^{- 8} \times (3000)^{4}$ . Since $300 0^{4} = 8.1 \times 1 0^{13}$ , this is $4.135 \times 1 0^{6} W/m^{2}$ . Why this step? The char glows white-hot; a large chunk of incoming heat leaves as light before it can ablate mass.
Net into ablation: $\overset{q}{˙}_{conv} - ε σ T_{w}^{4} = 8.0 \times 1 0^{6} - 4.135 \times 1 0^{6} = 3.865 \times 1 0^{6} W/m^{2}$ . Why this step? Only heat that is not re-radiated must be carried off by dying material.
Recession: $\overset{s}{˙} = \frac{3.865 \times 1 0 ^{6}}{270 \times 1.2 \times 1 0 ^{7}} = \frac{3.865 \times 1 0 ^{6}}{3.24 \times 1 0 ^{9}} = 1.193 \times 1 0^{- 3} m/s$ . Why this step? Divide net heat by (energy per m³ of shield) $= ρ_{abl} Q^{*}$ to get how many metres per second recede.

Answer: $\overset{s}{˙} \approx 1.19 \times 1 0^{- 3} m/s \approx 1.19 mm/s$ .

Verify: Re-radiation ate $4.1/8.0 \approx 52%$ of the incoming heat — over half — which is why hot char is a genuine third defence, not a footnote. Units: $(W/m^{2}) / [(kg/m^{3}) (J/kg)] = (J s^{- 1} m^{- 2}) / (J m^{- 3}) = m/s$ . ✓

Worked example Ex 7 — C7 Degenerate ablation: recession stalls to zero

Same PICA as Ex 6, but heating has dropped late in reentry to $\overset{q}{˙}_{conv} = 4.0 \times 1 0^{6} W/m^{2}$ while the char is still glowing at $T_{w} = 3000 K$ . Find $\overset{s}{˙}$ .

Forecast: From Ex 6 the re-radiation term is $4.135 MW/m^{2}$ . Incoming is $4.0 MW/m^{2}$ . What happens when the shield radiates more than arrives?

Net into ablation: $4.0 \times 1 0^{6} - 4.135 \times 1 0^{6} = - 1.35 \times 1 0^{5} W/m^{2}$ (negative). Why this step? The bracket in Tool 3 can go negative — this is the degenerate case the parent never showed.
Interpret the negative: a negative "recession" is unphysical — mass cannot un-ablate. The physical outcome is $\overset{s}{˙} = 0$ ; the surface stops receding and instead cools (its actual $T_{w}$ falls until radiation balances the reduced input). Why this step? We clamp at zero because the formula assumed steady ablation, which no longer holds when re-radiation alone can dump all incoming heat.

Answer: $\overset{s}{˙} = 0$ (ablation ceases; the shield now behaves like a re-radiating tile).

Verify: This is the physical meaning of $\overset{q}{˙}_{conv} \leq ε σ T_{w}^{4}$ : below that threshold no material is consumed. It also reinforces the parent's steel-man mistake "reusable tiles are always better than ablators" — the parent's fix was that at high flux no reusable solid survives, so ablation is required; this example shows the flip side, that at low flux even an ablator stops ablating and behaves reusable-like. Together they prove TPS choice is governed by the flux peak, not the average.

Worked example Ex 8 — C8 Real-world word problem: Mars shield mass

A Mars entry vehicle (see Mars Entry Descent and Landing) uses an SLA-561V shield, $ρ_{abl} = 260 kg/m^{3}$ . Engineers require a shield thickness of $5.0 cm$ over a circular heatshield of diameter $4.5 m$ . What mass of ablator is that?

Forecast: A 4.5 m disc, 5 cm thick, at roughly quarter-water density. Guess: tens of kg? Hundreds?

Shield frontal area: $A = π r^{2} = π (2.25)^{2} = 15.90 m^{2}$ . Why this step? Ablator is a layer on the front face; its volume is area × thickness.
Volume: $V = A \times t = 15.90 \times 0.05 = 0.7952 m^{3}$ . Why this step? Convert the 5 cm to $0.05$ m first so all lengths are SI.
Mass: $m = ρ_{abl} V = 260 \times 0.7952 = 206.8 kg$ . Why this step? Mass = density × volume — the definition of density rearranged.

Answer: $\approx 207 kg$ of ablator.

Verify: For a lander of a few tonnes, ~200 kg of heatshield is a believable single-digit-percent mass fraction — matching real EDL budgets. Units: $(kg/m^{3}) (m^{2}) (m) = kg$ . ✓

Worked example Ex 9 — C9 Exam twist: which TPS, and prove it with numbers

A wing leading edge sees $\overset{q}{˙}_{conv} = 1.5 \times 1 0^{6} W/m^{2}$ . A candidate reusable ceramic radiates with $ε = 0.8$ but melts at 2000 K. Using pure re-radiation, decide whether the ceramic survives; if not, name the TPS class that must be used instead and why.

Forecast: Compute the radiative-equilibrium $T_{w}$ and compare to 2000 K. Above → the material melts.

Equilibrium temp: $T_{w} = (\frac{q ˙ _{conv}}{ε σ})^{1/4} = (\frac{1.5 \times 1 0 ^{6}}{0.8 \times 5.67 \times 1 0 ^{- 8}})^{1/4}$ . Why this step? Same balance as Ex 5 — a reusable material can only survive if its equilibrium temperature stays below its melting point.
Inside the bracket: $\frac{1.5 \times 1 0 ^{6}}{4.536 \times 1 0 ^{- 8}} = 3.307 \times 1 0^{13}$ . Fourth root: $T_{w} = 2398 K$ . Why this step? Invert the $T^{4}$ with a fourth root, then substitute numbers to get a concrete kelvin value to compare against the 2000 K melting point.
Compare: $2398 K > 2000 K$ → the ceramic melts. Therefore a reusable radiator cannot survive this spot. You must use either RCC (carbon–carbon, rated >1900 K — the Shuttle nose/leading-edge material) or an ablator, whose $\overset{m}{˙} Q^{*}$ blocking term removes energy no solid could re-radiate. Why this step? This is the parent's steel-man mistake made quantitative: at high flux, $(\overset{q}{˙} / ε σ)^{1/4}$ can exceed every melting point, forcing ablation or the highest-temperature reusable material.

Answer: $T_{w} \approx 2398 K$ , above 2000 K → ceramic fails; choose RCC or an ablator.

Verify: This is exactly the Space Shuttle Columbia Accident lesson — leading edges get RCC precisely because ordinary tiles' equilibrium temperature blows past their limit. The number 2398 K > 2000 K settles the decision with no hand-waving.

Recall Self-test the matrix

Which cell needs a fourth-root? ::: C5 and C9 — inverting $ε σ T_{w}^{4}$ to get $T_{w}$ . Which cell makes $\overset{q}{˙} \to \infty$ ? ::: C2, as $R_{n} \to 0$ (sharp nose). Which cell makes $\overset{s}{˙}$ clamp to zero? ::: C7, when re-radiation ≥ incoming heat. Doubling speed multiplies heating by? ::: $2^{3} = 8$ (the $v^{3}$ law, cell C4). Above what speed does Tool 1 stop being enough? ::: ~ $10 km/s$ (Mach >25), where glowing-gas radiative input adds on top.