Thermal protection — silica tiles (Shuttle), UHTCs (ZrB₂, HfB₂): Worked examples

Before any symbol appears, let me re-earn each one in plain words, because you must never meet a letter you have not been introduced to.

Definition The symbols, in words and pictures

$q$ — heat flux: how much heat energy crosses one square metre of surface every second. Units: watts per square metre, $W/ m^{2}$ . Picture: arrows of heat poking through a window pane.
$k$ — thermal conductivity: how easily a material lets heat walk through it. Big $k$ = heat sprints through (metal spoon in tea). Small $k$ = heat crawls (a foam cup). Units $W m^{- 1} K^{- 1}$ .
$L$ — thickness of the slab the heat must cross, in metres. Picture: the width of the wall.
$Δ T$ (read "delta T") — the temperature difference between the hot side and the cold side, $Δ T = T_{hot} - T_{cold}$ . The triangle $Δ$ just means "the change in".
$ε$ (epsilon), $σ$ (sigma) — emissivity (0 to 1, how good a glower the surface is) and the Stefan–Boltzmann constant $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ , a fixed number of nature.
$V$ — re-entry velocity: how fast the vehicle plunges through the air, in metres per second (or km/s). Picture: the speedometer of the capsule as it hits the atmosphere. The parent note showed the peak heat flux grows as $q \propto V^{3}$ — speed is the cruellest term.
$ρ$ (read "rho") — air density: how much mass of air sits in each cubic metre, in $kg/ m^{3}$ . Picture: how "thick" the air is — thin high up, thicker as you descend. It enters the heating law as $q \propto ρ^{1/2}$ .
$R_{n}$ — nose radius: how blunt or sharp the leading edge is, in metres. A big $R_{n}$ is a rounded, blunt nose; a small $R_{n}$ is a needle-sharp edge. Picture: the radius of the circle that just fits the tip. The parent note showed peak heat flux scales as $q \propto R_{n}^{- 1/2}$ , so sharp = hot.

The scenario matrix

Every problem in this topic lands in exactly one of these cells. The examples below are labelled with their cell.

#	Cell (case class)	The twist it tests	Example
A	Conduction, normal sign ( $T_{hot} > T_{cold}$ )	forward use of $q = k Δ T / L$	Ex 1
B	Conduction, sign flip (which face is hot?)	negative $Δ T$ / direction of heat flow	Ex 2
C	Degenerate limit ( $L \to 0$ , sharp edge)	limit that breaks the tile philosophy	Ex 3 (figure)
C′	Degenerate limit ( $L \to \infty$ , thick block)	the twin limit, $q \to 0$	Ex 3b
D	Radiation forward ( $q \to T$ )	invert $q = ε σ T^{4}$	Ex 4
E	Radiation limit ( $ε \to 1$ vs small)	how emissivity shifts equilibrium $T$	Ex 5 (figure)
F	Velocity + geometry scaling (ratio $V^{3}$ , $R_{n}^{- 1/2}$ )	pure scaling, real word problem	Ex 6, Ex 6b
G	Zero / trivial input ( $q = 0$ , $Δ T = 0$ )	sanity of the degenerate answer	Ex 7
H	Combined / exam twist (conduction and radiation balance)	two laws at once, pick the philosophy	Ex 8 (figure)

The examples

Worked example Ex 1 — Cell A: conduction, normal sign

A silica tile has $k = 0.05 W m^{- 1} K^{- 1}$ , thickness $L = 0.05 m$ . The outer face sits at $T_{hot} = 1500 K$ and we want the back face held at $T_{cold} = 450 K$ . What steady conducted flux $q$ does this correspond to?

Forecast: low $k$ and a big temperature drop — do you expect a large or small $q$ ? Guess a rough magnitude before reading on.

Write Fourier's law in the steady, one-dimensional form: $q = k \frac{Δ T}{L}$ . Why this step? In steady state with no heat source inside, $q$ is the same at every depth, so the temperature falls in a straight line and $d T / d x = Δ T / L$ . That converts a calculus law into simple arithmetic.
Compute the drop: $Δ T = 1500 - 450 = 1050 K$ . Why this step? $Δ T$ is the driving force; heat only flows because the two faces disagree on temperature.
Plug in: $q = 0.05 \times \frac{1050}{0.05} = 1050 W/ m^{2}$ . Why this step? The two $0.05$ 's cancel — a happy coincidence of numbers — leaving $q = Δ T$ numerically here.

Verify: Units: $W m^{- 1} K^{- 1} \cdot K / m = W/ m^{2}$ ✓. Only $\sim 1$ kW/m² leaks through — tiny compared to the megawatt-scale surface flux, which is exactly why the tile insulates. $q = 1050 W/ m^{2}$ .

Worked example Ex 2 — Cell B: sign flip, which way does heat flow?

After landing, the same tile has cooled unevenly: the inner aluminium face is now at $600 K$ (still warm from soak-through) and the outer face is at $300 K$ (ambient). Same $k = 0.05$ , $L = 0.05 m$ . Which way does heat flow, and what is its magnitude?

Forecast: Heat always flows from hot to cold — so which face is losing heat now? Guess the direction before computing.

Identify hot and cold: now $T_{inner} = 600 > T_{outer} = 300$ . Heat flows inner → outer — the reverse of re-entry. Why this step? The formula does not know "front" from "back"; only the sign of $T_{hot} - T_{cold}$ picks the direction. Getting this wrong is the classic sign error.
Take the magnitude of the drop: $∣Δ T ∣ = 600 - 300 = 300 K$ . Why this step? We report flux magnitude; the direction we've already named in words, so we use the positive difference.
$q = 0.05 \times \frac{300}{0.05} = 300 W/ m^{2}$ , directed outward. Why this step? Same law, new numbers — the point was the reversed direction, not new physics.

Verify: If we had blindly written $Δ T = T_{outer} - T_{inner} = - 300$ , we'd get $q = - 300$ — the minus sign is the formula telling us the flow is opposite to our assumed direction. Consistent. $q = 300 W/ m^{2}$ inward-to-outward.

Worked example Ex 3 — Cell C: degenerate limit, the sharp edge (

$L \to 0$ ) A designer wants an aerodynamically perfect sharp leading edge, so the effective insulating thickness shrinks: $L = 0.05 m \to 0.005 m \to 0.0005 m$ , holding the same $Δ T = 1050 K$ and $k = 0.05$ . What happens to the conducted $q$ ?

Forecast: thinner wall — does the heat that reaches the structure go up or down? By how much as $L \to 0$ ?

Tabulate $q = k Δ T / L$ for the three thicknesses. Why this step? Watching the trend is the whole lesson: a single number hides the limit behaviour.
- $L = 0.05$ : $q = 0.05 \cdot 1050/0.05 = 1050 W/ m^{2}$
- $L = 0.005$ : $q = 0.05 \cdot 1050/0.005 = 10, 500 W/ m^{2}$
- $L = 0.0005$ : $q = 0.05 \cdot 1050/0.0005 = 105, 000 W/ m^{2}$
Read the limit: as $L \to 0$ , $q \to \infty$ . Why this step? $L$ sits in the denominator, so shrinking it blows $q$ up without bound.

What the figure shows: the horizontal axis is depth $x$ into the tile (metres), the vertical axis is temperature $T$ (kelvin). Each coloured line is a straight temperature profile from the 1500 K hot face down to the 450 K back face. As $L$ shrinks (blue → yellow → red), the same 1050 K drop is squeezed into a narrower slab, so the line tilts steeper and steeper — and the slope $Δ T / L$ is proportional to $q$ . The red near-vertical line is the sharp-edge blow-up: enormous slope, enormous flux.

Why this matters: A sharp edge has no thickness to insulate with. The "block the heat" tile philosophy literally diverges here — this is the mathematical reason UHTCs ("take the heat") exist for sharp edges. Ex 6b below turns this into a number using the $q \propto R_{n}^{- 1/2}$ law.

Verify: Each tenfold cut in $L$ multiplies $q$ by exactly $10$ ✓ (inverse proportionality). $q$ values: $1050, 10, 500, 105, 000 W/ m^{2}$ .

Worked example Ex 3b — Cell C′: the twin limit, a thick block (

$L \to \infty$ ) Now push the opposite way. Keep $k = 0.05$ and $Δ T = 1050 K$ , but make the tile ever thicker: $L = 0.5 m \to 5 m \to 50 m$ . What happens to the conducted $q$ ?

Forecast: If a sharp edge ( $L \to 0$ ) gave infinite leak, what should an infinitely thick wall give — a lot, a little, or nothing?

Reuse $q = k Δ T / L$ for the three thicknesses. Why this step? Same law, opposite corner of the input space — we're mapping the whole range of $L$ , not just one end.
- $L = 0.5$ : $q = 0.05 \cdot 1050/0.5 = 105 W/ m^{2}$
- $L = 5$ : $q = 0.05 \cdot 1050/5 = 10.5 W/ m^{2}$
- $L = 50$ : $q = 0.05 \cdot 1050/50 = 1.05 W/ m^{2}$
Read the limit: as $L \to \infty$ , $q \to 0$ . Why this step? $L$ in the denominator now grows without bound, so the fraction is driven to zero — a perfectly thick wall leaks nothing.

Why this matters: This is the mirror image of Ex 3. Together they bracket the entire conduction story: thickness is your insulation budget — spend a lot ( $L \to \infty$ , $q \to 0$ ) and nothing gets through; spend nothing ( $L \to 0$ , $q \to \infty$ ) and you're cooked. Real tiles sit in between, and the Shuttle chose $L \sim 0.05$ m as the sweet spot of mass vs protection.

Verify: Each tenfold increase in $L$ divides $q$ by $10$ ✓. $q$ values: $105, 10.5, 1.05 W/ m^{2}$ , heading to $0$ .

Worked example Ex 4 — Cell D: radiation forward, invert

$q = ε σ T^{4}$ A HfB₂ nose tip is in radiative equilibrium, dumping $q = 2.5 \times 1 0^{6} W/ m^{2}$ with emissivity $ε = 0.8$ . Find the surface temperature $T$ .

Forecast: With megawatts to shed and $T^{4}$ inside, will $T$ be near 1000 K, 2000 K, or 3000 K? Guess.

Set incoming = radiated: $q = ε σ T^{4}$ . Why this step? At steady state the only escape route for heat on a leading edge is glowing (radiation); there's no cool material behind to conduct into. So balance is legitimate.
Solve for $T$ by undoing the fourth power — take the fourth root: $T = (\frac{q}{ε σ})^{1/4} .$ Why the fourth root? $T$ was raised to the 4th; the 4th root is the operation that undoes it, just like a square root undoes a square.
Plug in: $\frac{2.5 \times 1 0 ^{6}}{0.8 \times 5.67 \times 1 0 ^{- 8}} = 5.51 \times 1 0^{13}$ , then $T = (5.51 \times 1 0^{13})^{1/4} \approx 2724 K$ . Why this step? Order of operations — do the division inside first, then one fourth-root at the end.

Verify: Reinsert: $0.8 \times 5.67 \times 1 0^{- 8} \times 272 4^{4} \approx 2.5 \times 1 0^{6}$ ✓. $T \approx 2724 K (\approx 2451^{\circ} C)$ — above silica's $\sim 1700^{\circ} C$ limit, so this must be a UHTC edge.

Worked example Ex 5 — Cell E: radiation limit, how much does

$ε$ move $T$ ? Same incoming flux $q = 1.0 \times 1 0^{6} W/ m^{2}$ . Compare the equilibrium temperature for a dull surface $ε = 0.3$ versus a black-glazed surface $ε = 0.95$ .

Forecast: A better glower ( $ε \to 1$ ) — does it run hotter or cooler for the same heat load? By a big factor or a small one?

Use $T = (q / (ε σ))^{1/4}$ twice. Why this step? Only $ε$ changes, so we isolate its effect.
- $ε = 0.30$ : $T = (\frac{1 0 ^{6}}{0.30 \times 5.67 \times 1 0 ^{- 8}})^{1/4} \approx 2560 K$
- $ε = 0.95$ : $T = (\frac{1 0 ^{6}}{0.95 \times 5.67 \times 1 0 ^{- 8}})^{1/4} \approx 1921 K$
Take the ratio to see the scaling of $ε$ : $\frac{T _{0.3}}{T _{0.95}} = (\frac{0.95}{0.30})^{1/4} \approx 1.334$ . Why this step? Because $ε$ sits under a fourth root, a 3.17× change in emissivity only shifts $T$ by 33% — the fourth root tames it.

What the figure shows: the horizontal axis is emissivity $ε$ (from 0.1 to 1.0), the vertical axis is the equilibrium surface temperature $T$ (kelvin). The blue curve is $T = (q / ε σ)^{1/4}$ : it falls as $ε$ rises, but gently — a whole sweep of $ε$ barely halves $T$ . The yellow dot marks the dull surface (0.30, ≈2560 K); the green dot marks the black-glazed surface (0.95, ≈1921 K). The near-flatness is the fourth root at work: emissivity is a weak lever on temperature.

Verify: $2560/1921 = 1.333$ ✓ and $(0.95/0.30)^{1/4} = 1.334$ ✓. Higher emissivity ⇒ cooler surface ⇒ this is why the Shuttle glaze is black ( $ε \approx 0.9$ ). $T_{0.3} \approx 2560$ K, $T_{0.95} \approx 1921$ K.

Worked example Ex 6 — Cell F: velocity scaling, a real word problem

A capsule returning from a Mars mission enters Earth's atmosphere at $V_{Mars} = 12.5 km/s$ ; a Soyuz from low orbit enters at $V_{Soyuz} = 7.8 km/s$ . Same nose radius $R_{n}$ and comparable air density $ρ$ . How many times worse is the Mars-return peak heating?

Forecast: The parent note says $q \propto V^{3}$ . Is the answer near 1.6×, 4×, or 40×?

Write the full scaling and form the ratio: $q \propto ρ^{1/2} V^{3} R_{n}^{- 1/2}$ , so $\frac{q _{Mars}}{q _{Soyuz}} = (\frac{ρ _{Mars}}{ρ _{Soyuz}})^{1/2} (\frac{V _{Mars}}{V _{Soyuz}})^{3} (\frac{R _{n, Soyuz}}{R _{n, Mars}})^{1/2}$ . Why this step? Taking a ratio is the trick: since $ρ$ and $R_{n}$ are the same for both ( $ρ_{Mars} = ρ_{Soyuz}$ , $R_{n, Mars} = R_{n, Soyuz}$ ), those two brackets each equal $1$ and cancel, leaving only the velocity term.
Compute the speed ratio: $\frac{12.5}{7.8} = 1.603$ . Why this step? Cube the ratio, not the speeds individually — cleaner and avoids huge numbers.
Cube it: $1.60 3^{3} \approx 4.12$ . Why this step? The exponent 3 comes from mass-flux $ρ V$ times kinetic-energy-per-mass $V^{2}$ , as derived in the parent note — since only $V$ differs between the two vehicles, the whole ratio collapses to $(V_{Mars} / V_{Soyuz})^{3}$ .

Verify: $1.60 3^{3} = 4.12$ ✓. About 4× worse — far beyond reusable tiles, which is why interplanetary returns use ablative shields (Ablative Heat Shields (Apollo, PICA)). $\approx 4.1 \times$ .

Worked example Ex 6b — Cell F: geometry scaling, sharp vs blunt (

$R_{n}^{- 1/2}$ ) Two nose tips re-enter at the same speed and air density, but one is blunt with $R_{n, blunt} = 0.30 m$ and the other is razor-sharp with $R_{n, sharp} = 0.003 m$ (100× smaller). How much hotter is the sharp tip's peak flux?

Forecast: The parent note says $q \propto R_{n}^{- 1/2}$ . A 100× smaller radius — does that give 10×, 100×, or 10000× the flux?

Form the ratio, cancelling the shared $ρ$ and $V$ : $\frac{q _{sharp}}{q _{blunt}} = (\frac{R _{n, blunt}}{R _{n, sharp}})^{1/2}$ . Why this step? Same ratio trick — everything identical except $R_{n}$ divides out, leaving the geometry term alone. Note $R_{n}^{- 1/2}$ means small $R_{n}$ goes upstairs, so sharp = big $q$ .
Plug in the radius ratio: $\frac{0.30}{0.003} = 100$ . Why this step? We need the ratio of radii before taking the root.
Take the square root (that's what the $1/2$ power means — it undoes a square): $100 = 10$ . Why this step? The exponent is $- 1/2$ , so a factor-100 shrink in $R_{n}$ becomes a factor- $100 = 10$ rise in $q$ .

Verify: $(0.30/0.003)^{1/2} = 10 0^{1/2} = 10$ ✓. The sharp edge sees 10× the heat flux of the blunt one — this is the quantitative version of "sharp = hot" that made Ex 3's tile philosophy diverge, and exactly why sharp edges demand UHTCs. $\approx 10 \times$ .

Worked example Ex 7 — Cell G: zero / trivial input (a sanity anchor)

Two degenerate checks: (a) the vehicle is on the pad, both tile faces at $300 K$ , so $Δ T = 0$ . (b) A perfect insulator, $k = 0$ . What is the conducted flux $q$ in each?

Forecast: No temperature difference, or no conductivity — should any heat cross? Guess before the algebra confirms.

Case (a): $q = k \frac{Δ T}{L} = 0.05 \times \frac{0}{0.05} = 0 W/ m^{2}$ . Why this step? If both faces agree on temperature there is no driving force, so heat has no reason to move — the formula must return zero, and it does.
Case (b): $q = 0 \times \frac{1050}{0.05} = 0 W/ m^{2}$ . Why this step? A material that conducts nothing ( $k = 0$ ) lets nothing through no matter how big the temperature drop — the ideal insulator. Confirms the formula behaves sanely at its edges.

Verify: Both give exactly $0$ ✓. These trivial cases bracket every real answer: real tiles sit between "perfect insulator" ( $q = 0$ ) and "no gradient" ( $q = 0$ ) with a small positive leak. $q = 0$ both.

Worked example Ex 8 — Cell H: exam twist, conduction AND radiation at once

A leading edge receives $q_{in} = 1.0 \times 1 0^{6} W/ m^{2}$ from the shock gas. The surface radiates away $q_{rad} = ε σ T_{s}^{4}$ with $ε = 0.85$ , and whatever is left conducts inward through a UHTC skin of $k = 60 W m^{- 1} K^{- 1}$ , $L = 0.01 m$ , to an interior held at $T_{cold} = 1000 K$ . Find the true equilibrium surface temperature $T_{s}$ .

Forecast: At steady state, in = out, but "out" now has TWO routes (radiate + conduct). Do you expect $T_{s}$ above or below the radiation-only estimate of Ex 4/5?

Write the balance: incoming = radiated + conducted, i.e. $q_{in} = ε σ T_{s}^{4} + k \frac{T _{s} - T _{cold}}{L}$ . Why this step? Energy can't pile up in steady state, so everything arriving must leave — and here there are two exits. This is the combined law the exam wants.
Try a guess, $T_{s} = 2100 K$ , to feel the balance before solving it exactly. Why this step? Testing a trial value shows which exit dominates. Radiated: $0.85 \times 5.67 \times 1 0^{- 8} \times 210 0^{4} = 9.37 \times 1 0^{5}$ . Conducted: $60 \times \frac{2100 - 1000}{0.01} = 6.60 \times 1 0^{6}$ . Their sum $7.54 \times 1 0^{6}$ far exceeds the $1.0 \times 1 0^{6}$ incoming — so $2100$ K is too hot; the true $T_{s}$ must be lower.
Solve properly. Because the high- $k$ conduction term is roughly linear and huge, it dominates the balance, so drop the small radiation term as a first pass: $q_{in} \approx k \frac{T _{s} - T _{cold}}{L}$ gives $T_{s} \approx T_{cold} + \frac{q _{in} L}{k} = 1000 + \frac{1 0 ^{6} \times 0.01}{60} = 1000 + 167 = 1167 K$ . Why this step? With $k = 60$ the conduction path is a wide-open drain; almost all heat leaves that way, pinning $T_{s}$ close to the backing temperature.
Check the dropped term at $T_{s} = 1167$ K: $q_{rad} = 0.85 \times 5.67 \times 1 0^{- 8} \times 116 7^{4} \approx 8.9 \times 1 0^{4}$ , about $9%$ of $q_{in}$ . Nudge $T_{s}$ down so radiation + conduction sum to exactly $1 0^{6}$ : solving $0.85 σ T_{s}^{4} + 6000 (T_{s} - 1000) = 1 0^{6}$ numerically gives $T_{s} \approx 1152 K$ . Why this step? The small radiation term shaves a few tens of degrees off the conduction-only estimate — including it makes the balance exact.

What the figure shows: the UHTC skin as a vertical blue bar. A red arrow ( $q_{in}$ ) drives heat in from the shock gas; a yellow arrow ( $q_{rad}$ ) glows heat back to space; a green arrow ( $q_{cond}$ ) drains heat inward to the 1000 K interior. The caption states the balance $q_{in} = q_{rad} + q_{cond}$ — the whole point is that both exits are open.

Verify: Plug $T_{s} = 1152$ K back: $q_{rad} = 0.85 \times 5.67 \times 1 0^{- 8} \times 115 2^{4} \approx 8.5 \times 1 0^{4}$ and $q_{cond} = 6000 \times (1152 - 1000) \approx 9.1 \times 1 0^{5}$ ; sum $\approx 9.96 \times 1 0^{5} \approx 1 0^{6}$ ✓. This closes Cell H: for a genuinely high- $k$ UHTC with a cool backing, conduction dominates and the surface runs far cooler (≈1152 K) than the radiation-only formula of Ex 4/5 would predict — you must include the conduction term.

Active recall

Recall Cover the answers — one per matrix cell

Which sign of $Δ T$ tells you heat flows inward? ::: When $T_{outer} > T_{inner}$ , i.e. $Δ T = T_{hot} - T_{cold} > 0$ with the hot face outside (Cell B). As $L \to 0$ (sharp edge), what does conducted $q$ do and why? ::: $q \to \infty$ because $L$ is in the denominator — the tile philosophy breaks, hence UHTCs (Cell C). As $L \to \infty$ (thick block), what does $q$ do? ::: $q \to 0$ — an infinitely thick wall leaks nothing; the twin limit of the sharp edge (Cell C′). How do you invert $q = ε σ T^{4}$ for $T$ ? ::: $T = (q / ε σ)^{1/4}$ — take the fourth root to undo the fourth power (Cell D). Why does raising emissivity from 0.3 to 0.95 only cool the surface by ~33%? ::: Because $ε$ sits under a fourth root, so even a 3× change moves $T$ only by $(3.17)^{1/4} \approx 1.33$ (Cell E). A 12.5 vs 7.8 km/s return is how much worse? ::: $(12.5/7.8)^{3} \approx 4.1 \times$ — only $V^{3}$ matters when $ρ, R_{n}$ match (Cell F). A 100× sharper nose is how much hotter? ::: $(100)^{1/2} = 10 \times$ — the $R_{n}^{- 1/2}$ geometry term (Cell F, Ex 6b). Why can't you use the radiation-only $T$ formula for a high- $k$ UHTC edge? ::: Because conduction carries a big share too; you must balance radiation and conduction, giving a much cooler $T_{s} \approx 1152$ K (Cell H).