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

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Step 1 — What does "heat arriving at a surface" even mean?

WHAT. Before any formula, we need one clean idea: heat flux. Imagine a tiny window of shield, exactly one square metre, facing the hot gas. In one second some amount of energy (in joules) passes through that window into the shield.

WHY. Every symbol later is either "energy per second per square metre" or something that turns into it. If we don't nail this unit picture first, the balance equation is just letters. We measure things per square metre because a big shield and a small shield made of the same stuff behave the same locally — only the local flux matters.

PICTURE. Look at the figure. The red square is our one-square-metre window. The black arrows are energy pouring in. We call the energy-per-second-per-square-metre the heat flux and write it $\dot{q}$.

• $\dot{q}$ ::: energy per second per square metre hitting the wall, in ${\text{W/m}}^2$
• Why "per square metre"? ::: so the result doesn't depend on how big the shield is — only on the local conditions

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Step 2 — Where does that flux come from? (the incoming term)

WHAT. The flux arriving at the wall during reentry is set by the shock-heated gas. From Reentry Aerothermodynamics and the parent note, its convective part scales as ${\dot{q}}_{\text{conv}}=k\sqrt{\frac{\rho }{R_n}}{v}^3.$

WHY. We list this so the reader knows $\dot{q}$ is not a free number — it is huge (megawatts per square metre) and it is given by the flight conditions, not by the shield. The shield's job is to survive whatever this delivers. We will call the net heat trying to enter the wall ${\dot{q}}_{\text{net}}$; think of it as "the leftover flux still pointed inward after we account for what leaves."

PICTURE. The red arrow is the incoming ${\dot{q}}_{\text{net}}$ striking the wall. Each symbol under the formula is labelled with its job.

• Why $v^3$ and not $v^2$? ::: energy per unit mass scales as $v^2$, and the mass arriving scales as $v$; multiplied → $v^3$
• Bigger nose radius does what to heating? ::: lowers it, because $\dot{q}\propto 1/\sqrt{R_n}$

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Step 3 — The wall fights back: it re-radiates

WHAT. A hot surface does not just sit there — it glows. Any object at temperature $T_w$ (the "w" is for wall) throws energy back out as light and infrared. The amount is ${\dot{q}}_{\text{out}}=\epsilon \sigma T_w^4.$

WHY. We introduce this because not all incoming flux ends up destroying shield material — some of it leaves again as radiation. If we forgot this term we would badly overestimate how fast the shield erodes. This is the Stefan–Boltzmann law, borrowed from Radiative Heat Transfer and Stefan–Boltzmann Law.

PICTURE. Same square, but now black arrows leave the surface. The hotter the wall, the more it glows — and because of the fourth power, doubling the temperature radiates sixteen times more.

• Why must $T_w$ be in kelvin? ::: radiation depends on absolute temperature; $T^4$ is only meaningful from absolute zero
• Why paint the char black (high $\epsilon$)? ::: high $\epsilon$ means it radiates more away, so more heat leaves before reaching the structure

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Step 4 — What's left over must be swallowed by dying material

WHAT. Take the incoming flux, subtract what radiates away. Whatever is left is the energy that goes into physically destroying shield material — breaking chemical bonds, melting, boiling. Call the mass removed per second per square metre $\dot{m}$, and the energy needed to remove one kilogram $Q^{\ast }$. Then the energy the ablation soaks up per second per square metre is $\dot{m}{Q}^{\ast }$.

WHY. This is the heart of the ablator idea, "die to survive." Energy has to be conserved at the surface: everything coming in must be accounted for by (leaving as light) + (spent killing material). There is nowhere else for it to go in steady state.

PICTURE. The red layer is the shield material turning to gas and streaming away; the little arrows are the mass leaving. The energy it carries off is $\dot{m}{Q}^{\ast }$.

• Meaning of $Q^{\ast }$ ::: energy absorbed per kilogram of shield destroyed
• Why is a high $Q^{\ast }$ desirable? ::: each kilogram of shield swallows more heat, so you lose less material for the same heat load

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Step 5 — Write the balance and read it as an equation

WHAT. Put the three flows on one line. In goes ${\dot{q}}_{\text{net}}$; out goes radiation $\epsilon \sigma T_w^4$; the rest is ablation $\dot{m}{Q}^{\ast }$: ${\dot{q}}_{\text{net}}=\dot{m}{Q}^{\ast }+\epsilon \sigma T_w^4.$

WHY. This is just "books balance": energy in = energy accounted for. No new physics — only bookkeeping of the three arrows we drew in Steps 2, 3, 4. This single line is the parent note's boxed energy balance.

PICTURE. All three arrows on one square: red = incoming ${\dot{q}}_{\text{net}}$, black up = radiation, black away = ablation mass. The equation is the promise that the red equals the two blacks combined.

• What principle is this line? ::: conservation of energy at the surface — in equals out
• Which term disappears if the wall is cold? ::: the radiation term $\epsilon \sigma T_w^4$ shrinks toward zero as $T_w\to 0$

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Step 6 — Solve for what an engineer actually wants: recession rate

WHAT. An engineer doesn't measure $\dot{m}$ directly — they want to know how fast the surface recedes, i.e. how many millimetres of shield vanish per second. Call that $\dot{s}$ (s for surface position). Mass removed and thickness removed are linked by the material's density ${\rho }_{\text{abl}}$: $\dot{m}={\rho }_{\text{abl}}\dot{s}.$ Substitute into the balance and solve for $\dot{s}$:

\;\Longrightarrow\; \boxed{\;\dot s=\frac{\dot q_\text{net}-\varepsilon\sigma T_w^4}{\rho_\text{abl}\,Q^*}\;}$$ **WHY.** We divide by $\rho_\text{abl}Q^*$ to isolate $\dot s$ because that is the design number: multiply $\dot s$ by the flight time and you get how thick the shield must be so it doesn't burn all the way through. The subtraction in the numerator says the wall only starts losing mass with whatever flux *survives* the re-radiation. **PICTURE.** The red line is the retreating surface; the arrow $\dot s$ shows it marching backward into the shield at some millimetres per second. The old surface (dashed) is gone. > [!formula] Recession rate — the target result > $$\dot s=\frac{\dot q_\text{net}-\varepsilon\sigma T_w^4}{\rho_\text{abl}\,Q^*}$$ > - Numerator $\dot q_\text{net}-\varepsilon\sigma T_w^4$ — the *net* flux left after radiation; only this drives material loss. > - $\rho_\text{abl}$ — shield density; a lighter, low-density material (PICA at $270\ \text{kg/m}^3$) recedes *faster* per kilogram but weighs less overall. > - $Q^*$ — high value = slow recession. ![[deepdives/dd-physics-3.4.22-d2-s06.png]] - Why divide by $\rho_\text{abl}Q^*$? ::: to convert "energy left over" into "thickness lost per second," the quantity a designer sizes the shield with - What multiplies $\dot s$ to give required shield thickness? ::: the total reentry duration (heating time) --- ## Step 7 — Edge cases: where does the formula behave strangely? **WHAT.** A good formula must survive its extremes. Three cases: 1. **Cold start ($T_w\approx0$).** Radiation term $\varepsilon\sigma T_w^4\to0$, so $\dot s\to\dot q_\text{net}/(\rho_\text{abl}Q^*)$ — *maximum* recession. The shield erodes fastest at the instant of first heating before it warms up and starts glowing. 2. **Radiation-dominated ($\varepsilon\sigma T_w^4=\dot q_\text{net}$).** Numerator hits zero, so $\dot s=0$ — **no ablation at all**. All incoming heat leaves as light. This is exactly the *reusable-tile* limit from the parent note: a tile survives precisely when it can radiate everything away. See [[Radiative Heat Transfer and Stefan–Boltzmann Law]]. 3. **Over-radiating? ($\varepsilon\sigma T_w^4>\dot q_\text{net}$).** Numerator goes negative, $\dot s<0$. A negative recession is unphysical — you can't grow shield back. What it really means: the wall *cannot* stay that hot; it would radiate more than it receives, so it cools until balance returns. The equation self-corrects by driving $T_w$ down until the numerator is $\ge0$. **WHY.** These cases connect this one formula to the *entire* TPS decision table in the parent: case 2 is a tile, cases 1 and 3 bracket a real ablator's life. Nothing in the parent's story is outside this equation. **PICTURE.** Three mini-panels: (left) cold, fast recession; (middle) balanced, zero recession = the tile limit; (right) the impossible negative case with an arrow showing $T_w$ falling back to balance. ![[deepdives/dd-physics-3.4.22-d2-s07.png]] > [!mistake] "If $T_w$ is huge the shield recedes negative — the equation is broken." > **Why it feels right:** you can plug in a big $T_w$ and get $\dot s<0$. > **The fix:** $T_w$ is not a free knob. In steady state the wall temperature *settles* at whatever value makes the books balance with $\dot s\ge0$. A negative answer just tells you your assumed $T_w$ was too high; physically it drops until $\dot s=0$ (the tile case) or until ablation restarts. - Cold-wall limit gives what recession? ::: the maximum, $\dot s=\dot q_\text{net}/(\rho_\text{abl}Q^*)$, since radiation is negligible - Zero-recession condition ::: $\varepsilon\sigma T_w^4=\dot q_\text{net}$ — everything re-radiates, the reusable-tile limit --- ## Step 8 — A worked number to make it real **WHAT.** Take a PICA-like shield at a moderate reentry point: $\dot q_\text{net}=2.0\times10^{6}\ \text{W/m}^2$, wall glowing at $T_w=2000\ \text{K}$, $\varepsilon=0.9$, $\sigma=5.67\times10^{-8}$, density $\rho_\text{abl}=270\ \text{kg/m}^3$, and $Q^*=1.2\times10^{7}\ \text{J/kg}$. **WHY.** Numbers convert the algebra into "will my shield survive." We compute the radiated flux, the net driving flux, and finally $\dot s$. **Radiated flux:** $$\varepsilon\sigma T_w^4=0.9\times5.67\times10^{-8}\times(2000)^4=0.9\times5.67\times10^{-8}\times1.6\times10^{13}\approx8.16\times10^{5}\ \text{W/m}^2.$$ **Recession rate:** $$\dot s=\frac{2.0\times10^{6}-8.16\times10^{5}}{270\times1.2\times10^{7}}=\frac{1.184\times10^{6}}{3.24\times10^{9}}\approx3.65\times10^{-4}\ \text{m/s}.$$ That is about **0.37 mm per second**. Over a 120-second heat pulse, the shield loses roughly $0.37\times120\approx44\ \text{mm}$ — so a 6 cm shield survives with margin. Notice the glowing wall shed about **41\%** of the incoming flux by radiation before any material was lost. ![[deepdives/dd-physics-3.4.22-d2-s08.png]] - Recession rate for the worked case ::: about $3.65\times10^{-4}$ m/s, i.e. $\approx0.37$ mm/s - Fraction of incoming flux radiated away ::: $\approx0.816/2.0\approx41\%$ --- ## The one-picture summary ![[deepdives/dd-physics-3.4.22-d2-s09.png]] The whole derivation in one frame: incoming red flux $\dot q_\text{net}$ strikes the wall; part leaves as glow $\varepsilon\sigma T_w^4$; the remainder drives mass $\dot m\,Q^*$ off the surface; converting mass loss to thickness loss via $\rho_\text{abl}$ gives the recession rate $\dot s$. > [!recall]- Feynman retelling of the whole walkthrough > Picture a one-square-metre patch of heat shield. Reentry gas dumps a firehose of energy onto it — millions of watts per square metre (Step 1–2). The patch fights back two ways. First, once it's glowing hot it throws energy *back out* as light, and because glow grows with temperature to the fourth power, a bright char sheds a big chunk of the incoming heat (Step 3). Second, whatever heat is *still* left over rips the shield material apart — breaking bonds, boiling it off — and that gas streams away carrying energy with it; each kilogram carried off eats $Q^*$ joules (Step 4). Energy has to balance: what comes in equals what glows out plus what boils off (Step 5). Since engineers care about *how fast the shield shrinks*, we swap "mass per second" for "millimetres per second" using the material's density and solve — that's the recession rate $\dot s$ (Step 6). Check the extremes: a cold wall erodes fastest, a wall that can radiate everything doesn't erode at all (that's a reusable tile), and if you assume the wall is too hot the math says "impossible," which really means the wall cools until the books balance (Step 7). Plug in real PICA numbers and you find a shield retreating under half a millimetre per second — a few centimetres of material outlasts the whole fiery plunge (Step 8). > [!mnemonic] > **IN = GLOW + GONE.** Incoming heat equals what glows away plus what's gone (ablated). Divide "gone" by density to get how fast the surface shrinks. --- **See also:** [[Bow Shock and Stagnation Point]] · [[Blunt Body Aerodynamics]] · [[Space Shuttle Columbia Accident]] · [[Mars Entry Descent and Landing]] · [[Specific Impulse and Energy Budgets]]