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

Before we start, two things you will reuse everywhere.

Definition The stagnation-point heating law — and where its symbols come from

At the front of a re-entering vehicle, the incoming convective heat flux (power per unit area, in $W/ m^{2}$ ) follows $q \propto R_{n}^{- 1/2} ρ^{1/2} V^{3}$ Each symbol is a plain physical picture:

$R_{n}$ = nose radius — how rounded the front of the vehicle is (metres). A big blunt nose = large $R_{n}$ ; a knife-edge = tiny $R_{n}$ .
$ρ$ = atmospheric density — how thick the air is at that altitude ( $kg/ m^{3}$ ). High deep in the atmosphere, thin up high.
$V$ = the vehicle's speed relative to the air ( $m/s$ ).

Where it comes from (plain-words derivation): the air rushing into the shock delivers mass at a rate $\overset{m}{˙} \propto ρ V$ (density × speed), and each kilogram carries kinetic energy $\propto V^{2}$ . Multiply mass rate by energy-per-mass: power flux $\propto ρ V \cdot V^{2} = ρ V^{3}$ — that is the $ρ^{1/2} V^{3}$ family (the square-root on $ρ$ comes from boundary-layer thinning, below). The $R_{n}^{- 1/2}$ factor is from boundary-layer theory (Fay–Riddell): a smaller nose has a thinner layer of slowed air, which conducts heat to the wall faster — so sharp edges are punished. Full aerodynamic background: Re-entry Aerodynamics & Boundary Layers.

Intuition Convection vs radiation — which heats the edge?

Two different mechanisms deliver heat to a re-entry surface:

Convective heating — the shock-heated gas physically touches the wall and conducts energy into it. This is the $q \propto R_{n}^{- 1/2} ρ^{1/2} V^{3}$ law above. It dominates for orbital and most planetary re-entry (roughly $V ≲ 10$ – $11 km/s$ ).
Radiative heating — at very high speeds ( $V ≳ 11 km/s$ , e.g. lunar/interplanetary return) the shock gas itself glows so hot it radiates energy onto the wall as well.

In the exercises below, the incoming flux $q$ (however it was delivered) is given, and the surface re-emits by radiative cooling $q = ε σ T^{4}$ . Setting incoming = re-radiated finds the surface temperature. Keep the two roles separate: convection/radiation bring heat in; Stefan–Boltzmann takes it out.

Level 1 — Recognition

These test whether you can name and sort the ideas without calculation.

Recall Solution

(a) Block the heat — silica tiles. A broad flat surface has lots of area and a thick tile can sit behind it. The goal is to insulate the aluminium frame and re-radiate incoming heat back to space. Low thermal conductivity is the winning property here.

(b) Take the heat — UHTCs (ZrB₂ / HfB₂). A sharp edge has a tiny nose radius $R_{n}$ , so the heat flux is huge and there is no volume to insulate with. You must pick a material that is genuinely stable at the equilibrium temperature and conducts heat away. That is the UHTC job.

Recall Solution

Velocity $V$ . It enters as $V^{3}$ , so doubling it multiplies the flux by $2^{3} = 8$ . By comparison doubling $ρ$ gives $2^{1/2} \approx 1.41$ , and doubling $R_{n}$ reduces flux by $2^{- 1/2} \approx 0.71$ . Velocity is king — that is why lunar return is far worse than low-orbit return.

Level 2 — Application

Plug the standard formulas. Watch your units.

Recall Solution

From Fourier's law in steady 1-D form, $q = k Δ T / L$ , so $Δ T = \frac{q L}{k} = \frac{800 \times 0.05}{0.05} = 800 K .$ Why this step: in steady state with no heat generated inside the tile, the flux $q$ is the same at every depth, which forces a straight-line temperature profile — so the drop is simply flux × thickness ÷ conductivity. See Fourier's Law of Heat Conduction. An 800 K drop is exactly the insulating margin that keeps the aluminium below its ~175 °C limit.

Recall Solution

At equilibrium, incoming flux = radiated flux, so $q = ε σ T^{4}$ . Solve for $T$ : $T = (\frac{q}{ε σ})^{1/4} = (\frac{5.0 \times 1 0 ^{5}}{0.90 \times 5.67 \times 1 0 ^{- 8}})^{1/4} .$ The bracket is $\frac{5.0 \times 1 0 ^{5}}{5.103 \times 1 0 ^{- 8}} = 9.798 \times 1 0^{12}$ . Taking the fourth root: $T = (9.798 \times 1 0^{12})^{1/4} \approx 1768 K (\approx 1495^{\circ} C) .$ Why this step: at steady state the only escape route for the absorbed heat is radiation, so setting absorbed = radiated is the correct energy balance. This 1768 K sits right at silica's ~1700 °C (1973 K) ceiling — a warning that we are near the switch-over to UHTCs.

Level 3 — Analysis

Now compare, scale, and reason about why.

Recall Solution

Only the $V^{3}$ term differs, so the ratio is $\frac{q _{Mars}}{q _{LEO}} = (\frac{12.5}{7.5})^{3} = (1.6667)^{3} \approx 4.63.$ Why this step: because $ρ$ and $R_{n}$ cancel, the whole scaling collapses to the cube of the velocity ratio. A ~4.6× higher flux is exactly why interplanetary return demands an ablative shield (see PICA) rather than reusable tiles. (Above ~11 km/s the radiative heating from the glowing shock also switches on, making the real gap even larger — but the convective cube already tells the story.)

Recall Solution

First convert: $1700^{\circ} C = 1700 + 273 = 1973 K$ . $q = ε σ T^{4} = 0.85 \times 5.67 \times 1 0^{- 8} \times (1973)^{4} .$ Compute $(1973)^{4} = 1.516 \times 1 0^{13}$ . Then $q = 0.85 \times 5.67 \times 1 0^{- 8} \times 1.516 \times 1 0^{13} \approx 7.31 \times 1 0^{5} W/ m^{2} .$ Why this step: we inverted the equilibrium equation — instead of "given flux, find temperature," we asked "at what flux does the temperature hit the material's ceiling?" Any leading edge seeing more than about $7.3 \times 1 0^{5} W/ m^{2}$ would drive silica past its limit and force a UHTC choice.

Figure — Thermal protection — silica tiles (Shuttle), UHTCs (ZrB₂, HfB₂)

Level 4 — Synthesis

Combine multiple ideas into one decision.

Recall Solution

Zone A: $T_{A} = (\frac{1.2 \times 1 0 ^{5}}{0.85 \times 5.67 \times 1 0 ^{- 8}})^{1/4} = (2.4906 \times 1 0^{12})^{1/4} \approx 1256 K .$ $1256 K$ is comfortably below the $1973 K$ silica ceiling → silica tiles (block/insulate) are the right, lighter, reusable choice.

Zone B: $T_{B} = (\frac{1.5 \times 1 0 ^{6}}{0.85 \times 5.67 \times 1 0 ^{- 8}})^{1/4} = (3.1133 \times 1 0^{13})^{1/4} \approx 2363 K .$ $2363 K$ ( $\approx 2090^{\circ} C$ ) is far above silica's limit but well below ZrB₂'s $\approx 3520 K$ melting point → UHTC (ZrB₂ + SiC) is required: it stays solid, conducts heat away from the tip, and self-heals a protective borosilicate glass. Why this synthesis: one equation ( $q = ε σ T^{4}$ ) plus one threshold (silica ceiling) plus one bonding fact (UHTC $T_{m}$ ) fully determines the choice for each zone.

Recall Solution

Cause-and-effect chain:

ZrB₂ oxidises, producing $B_{2} O_{3}$ (a glass, melts $\approx 450^{\circ} C$ ) + porous $Zr O_{2}$ .
$B_{2} O_{3}$ has an appreciable vapour pressure that climbs steeply with temperature; above roughly 1100 °C it evaporates fast enough that the glassy layer is lost as quickly as it forms. (This ~1100 °C figure is the practical onset where $B_{2} O_{3}$ volatilisation outruns its replenishment — it is a kinetic boundary, not a fixed boiling point.)
Losing the glass leaves only porous $Zr O_{2}$ , through which oxygen diffuses freely → runaway oxidation.
Adding SiC makes $Si O_{2}$ , whose glass is far more viscous and far less volatile; it combines with residual $B_{2} O_{3}$ to form a borosilicate glass that stays put to well above 1600 °C.
Being viscous and continuous, this glass is a poor pathway for oxygen — the diffusion of $O_{2}$ through it is slow, so the underlying ceramic is shielded and the part survives. Key insight: the protection is kinetic (slow O₂ transport through a viscous, non-volatile glass), not thermodynamic inertness. See Oxidation Kinetics and Protective Oxide Layers.

Level 5 — Mastery

Open-ended reasoning across the whole chapter.

Recall Solution

Broad flat surface (silica): there is plenty of depth behind the skin. The goal is to keep the inner structure cold, so you want to impede heat flow through the thickness. Low $k$ ⇒ steep temperature drop across the tile ⇒ cold aluminium frame (Fourier: $Δ T = q L / k$ , small $k$ ⇒ big $Δ T$ ).

Sharp tip (UHTC): a knife-edge concentrates flux into a tiny volume with nowhere to insulate. If heat could not move sideways, that point would spike far above the bulk temperature and melt locally. High $k$ spreads the heat along the edge into cooler material, keeping any local hotspot below the limit.

Unifying idea: the design philosophy sets the requirement. "Block the heat" ⇒ minimise transport (low $k$ ). "Take the heat, avoid hotspots" ⇒ maximise lateral transport (high $k$ ). Same property, opposite optimum, because the geometry and the goal are opposite. This ties to Covalent vs Metallic Bonding: the metallic Zr sublattice in ZrB₂ is what gives UHTCs their high $k$ , whereas amorphous SiO₂'s random covalent network is what destroys conduction paths (see Amorphous vs Crystalline Solids).

Recall Solution

Allowed temperature: $T_{max} = \frac{3650}{1.5} \approx 2433 K$ . Maximum flux from the radiative balance: $q_{max} = ε σ T_{max}^{4} = 0.85 \times 5.67 \times 1 0^{- 8} \times (2433)^{4} .$ $(2433)^{4} = 3.504 \times 1 0^{13}$ , so $q_{max} = 0.85 \times 5.67 \times 1 0^{- 8} \times 3.504 \times 1 0^{13} \approx 1.69 \times 1 0^{6} W/ m^{2} .$ Why this step: a safety factor on absolute temperature translates through the fourth-power radiation law into a flux ceiling. Note how forgiving the fourth power is: a modest 1.5× temperature cushion still permits well over $1 0^{6} W/ m^{2}$ — this is the deep reason radiative cooling is so effective at extreme temperature.

Recall Master checklist (cover and recite)

Two philosophies? ::: block/insulate (silica) vs take/survive (UHTC).
Which term dominates $q$ ? ::: $V^{3}$ — velocity, cube scaling.
What are $R_{n}$ and $ρ$ ? ::: nose radius and atmospheric density.
Convective vs radiative heating in? ::: gas touches wall (convective, $\propto R_{n}^{- 1/2} ρ^{1/2} V^{3}$ ) vs glowing shock (radiative, high speed).
Silica back-face temperature? ::: $Δ T = q L / k$ (low $k$ ⇒ big drop).
Radiative equilibrium out? ::: $T = (q / ε σ)^{1/4}$ .
Why SiC in ZrB₂? ::: forms non-volatile borosilicate glass; kinetic O₂ barrier.
Conductivity: silica vs UHTC? ::: low $k$ to insulate vs high $k$ to spread heat.