Exercises — Ablative cooling — charring, blowing

What: We pick the definition. Answer: (b) ablative cooling. Why: "Passive" means the material does the work by itself — it decomposes and blows its own gas. Regenerative cooling (Regenerative Cooling) pumps cold propellant through channels; that needs plumbing and pressure — active. So it is not (a). Any pumped film cooling is active too, ruling out (c).

What: Trace the path from cold structure outward to hot gas.

• Virgin material → (ii) still cold & unreacted.
• Pyrolysis zone → (iii) resin decomposes, releases gas.
• Char layer → (i) porous carbon insulator.
• Boundary layer → (iv) gas cushion. Why this order: heat comes from the outside, so the hottest, most-transformed zone (char, then boundary layer) is on top; the cold intact resin is deepest.

Because breaking bonds, vaporizing and blowing gas each consume energy that would otherwise reach the structure, and the escaping gas physically blocks the hot gas from touching the surface. The heat is spent on destroying the surface instead of heating the interior.

What: Use $Q^{\ast }={\dot{q}}_{net}/{\dot{m}}^{\prime \prime }$, so ${\dot{m}}^{\prime \prime }={\dot{q}}_{net}/Q^{\ast }$. Why: $Q^{\ast }$ is "joules rejected per kg lost". Dividing the joules-per-second-per-area by joules-per-kg leaves kg-per-second-per-area — exactly ${\dot{m}}^{\prime \prime }$. Units force this: $\frac{{\text{W/m}}^2}{\text{J/kg}}=\frac{{\text{J/s⋅m}}^2}{\text{J/kg}}={\text{kg/m}}^2\text{s}$. ✓ ${\dot{m}}^{\prime \prime }=\frac{4\times 10^6}{1.0\times 10^7}=0.40{\text{kg/m}}^2\text{s}$ Meaning: every second, each square metre of surface loses 0.4 kg of material to handle that heat.

What (step 1): convert mass loss to a receding depth using $\dot{s}={\dot{m}}^{\prime \prime }/{\rho }_v$. Why: mass per area per second ÷ (mass per volume) = volume per area per second = length per second — the speed the surface eats inward. $\dot{s}=\frac{0.40}{1600}=2.5\times 10^{-4}\text{m/s}=0.25\text{mm/s}$ What (step 2): multiply by time (steady flux ⇒ constant speed). $s=\dot{s}\cdot t=2.5\times 10^{-4}\times 80=0.020\text{m}=20\text{mm}$ Meaning: you need a liner thicker than 2 cm to survive this burn with margin.

What: plug into $B^{\prime }=\frac{{\dot{m}}^{\prime \prime }{c}_p}{h_0}$. Why this grouping: ${\dot{m}}^{\prime \prime }{c}_p$ has units ${\text{W/m}}^2\text{K}$ (a "blowing conductance"); dividing by $h_0$ (same units) gives a pure number — how hard we blow compared to how hard the gas leaks heat in. $B^{\prime }=\frac{0.30\times 1200}{900}=\frac{360}{900}=0.40$ Meaning: $B^{\prime }=0.4$ is moderate blowing — enough to matter but far from blow-off.

What: use $\frac{h}{h_0}=\frac{\ln (1+B^{\prime })}{B^{\prime }}$. Why this formula, not something simpler: blowing does not subtract heat linearly — the injected gas thickens the boundary layer (Boundary Layer Theory), and solving the 1-D energy balance with wall injection produces exactly the logarithm. That log is why cooling saturates. $\frac{h}{h_0}=\frac{\ln (1.4)}{0.4}=\frac{0.33647}{0.4}=0.8412$ Meaning: heating is reduced to about 84% of the un-blown value, i.e. a cut of roughly $16\%$. Modest — because $B^{\prime }$ is small.

What: evaluate the factor at both points. $B^{\prime }=0.4:\frac{\ln 1.4}{0.4}=0.841B^{\prime }=4:\frac{\ln 5}{4}=\frac{1.6094}{4}=0.402$ Going from $B^{\prime }=0.4$ to $B^{\prime }=4$ (a $10\times$ increase in blowing) only drops $h/h_0$ from $0.84$ to $0.40$ — roughly halving it, not tenfold. Why: the curve in the figure is $\ln (1+B^{\prime })/B^{\prime }$, which flattens as $B^{\prime }$ grows (the red curve bends toward zero slowly). Physically, once the boundary layer is already thick, extra gas has less hot gas left to push away — diminishing returns. You just erode faster for little extra protection.

(a) $B^{\prime }\to 0$: use $\ln (1+B^{\prime })\approx B^{\prime }-\frac{B^{\prime 2}}{2}$ for small $B^{\prime }$. Then $\frac{\ln (1+B^{\prime })}{B^{\prime }}\approx \frac{B^{\prime }-B^{\prime 2}/2}{B^{\prime }}=1-\frac{B^{\prime }}{2}\to 1.$ So $h\to h_0$: no blowing ⇒ no reduction. Sanity check passes. (b) $B^{\prime }\to \infty$: the numerator $\ln (1+B^{\prime })$ grows like $\ln B^{\prime }$ (slow), the denominator like $B^{\prime }$ (fast), so the ratio $\to 0$. Heating shuts off — this is the blow-off limit. Meaning: the formula behaves correctly at both ends, which is exactly why we trust it in between.

What is the "blocking term"? It is the third slot of the master formula from the parent note, $Q^{\ast }=\underset{\text{pyrolysis}}{\underbrace{Q_p}}+\underset{\text{sensible}}{\underbrace{c_p(T_w-T_i)}}+\underset{\text{blocking by blowing}}{\underbrace{\eta \frac{h_0}{{\dot{m}}^{\prime \prime }}(H_{gas}-H_w)}}.$ The blocking term $\eta \frac{h_0}{{\dot{m}}^{\prime \prime }}(H_{gas}-H_w)$ is the un-blown convective load $h_0(H_{gas}-H_w)$ that blowing prevents from arriving, divided by the mass rate ${\dot{m}}^{\prime \prime }$ that did the blocking — i.e. joules blocked per kilogram lost — with an efficiency factor $\eta$. In this problem that whole group has already been evaluated for us as $5.5\times 10^6\text{J/kg}$; we take it as a given number. Why we add the three: each term measures energy per kg the material handles by a different mechanism (breaking bonds, warming up, blowing gas). They don't overlap, so they simply add.

• Sensible: $c_p(T_w-T_i)=1400\times (3000-300)=1400\times 2700=3.78\times 10^6\text{J/kg}$.
• Pyrolysis: $2.5\times 10^6\text{J/kg}$.
• Blocking: $5.5\times 10^6\text{J/kg}$. $Q^{\ast }=(2.5+3.78+5.5)\times 10^6=1.178\times 10^7\text{J/kg}.$ Dominant channel: the blocking/blowing term ($5.5\times 10^6$) is the largest — confirming blowing is a main cooling channel, not a side effect.

What (step 1): ${\dot{m}}^{\prime \prime }={\dot{q}}_{net}/Q^{\ast }$. ${\dot{m}}^{\prime \prime }=\frac{6\times 10^6}{1.178\times 10^7}=0.5093{\text{kg/m}}^2\text{s}$ What (step 2): total mass per area over the burn = ${\dot{m}}^{\prime \prime }\times t$. $m^{\prime \prime }=0.5093\times 120=61.1{\text{kg/m}}^2$ What (step 3): recession depth $s={\dot{m}}^{\prime \prime }t/{\rho }_v=m^{\prime \prime }/{\rho }_v$. $s=\frac{61.1}{1450}=0.04215\text{m}=42.2\text{mm}$ Meaning: the surface recedes about $42\text{mm}$, so a $30\text{mm}$ liner does not survive — it burns fully through. You would need thicker material or a higher-$Q^{\ast }$ ablator.

What: for each, ${\dot{m}}^{\prime \prime }=\dot{q}/Q^{\ast }$, then depth $s={\dot{m}}^{\prime \prime }t/{\rho }_v$, and consumed mass $m^{\prime \prime }={\dot{m}}^{\prime \prime }t$.

Material A: ${\dot{m}}_A^{\prime \prime }=\frac{8\times 10^6}{1.2\times 10^7}=0.6667{\text{kg/m}}^2\text{s}$ $m_A^{\prime \prime }=0.6667\times 90=60.0{\text{kg/m}}^2,s_A=\frac{60.0}{1450}=0.04138\text{m}=41.4\text{mm}$

Material B: ${\dot{m}}_B^{\prime \prime }=\frac{8\times 10^6}{0.7\times 10^7}=1.1429{\text{kg/m}}^2\text{s}$ $m_B^{\prime \prime }=1.1429\times 90=102.9{\text{kg/m}}^2,s_B=\frac{102.9}{550}=0.1870\text{m}=187\text{mm}$

Depth winner: Material A recedes only $41\text{mm}$ vs B's $187\text{mm}$ — A wins hugely on thickness needed (higher $Q^{\ast }$ + higher density). Consumed-mass winner: A loses $60{\text{kg/m}}^2$ vs B's $103{\text{kg/m}}^2$ — A also loses less mass, because its $Q^{\ast }$ is larger (more joules per kg). The trade-off: A wins both raw metrics here, BUT B is far less dense ($550$ vs $1450$). If the mission is mass-limited, a thick B liner might still weigh less per unit area for a given standoff, or B may be cheaper/easier to shape. Real selection weighs $Q^{\ast }$, density, char stability (Re-entry Aerothermodynamics loads), and manufacturability — not depth alone.

What / the loop (follow Figure s02 clockwise):

1. More heat arrives → more pyrolysis → larger ${\dot{m}}^{\prime \prime }$.
2. Larger ${\dot{m}}^{\prime \prime }$ → larger blowing parameter $B^{\prime }={\dot{m}}^{\prime \prime }{c}_p/h_0$.
3. Larger $B^{\prime }$ → smaller $h/h_0=\ln (1+B^{\prime })/B^{\prime }$ → less heat arrives (the red arrow closing the loop). So step 3 fights step 1 — a negative feedback that stabilises the surface: the ablator blows harder exactly when it is hit harder. Why enthalpy $H$, not temperature $T$: at re-entry temperatures the gas dissociates and recombines (Convective Heat Transfer (Stanton number)). The true driving potential for heat transfer is the enthalpy gap $H_{gas}-H_{wall}$, because chemical energy of recombination also deposits at the wall — using $T$ alone omits that chemical energy and therefore underestimates the true heat load. Enthalpy bundles thermal and chemical energy into one number, so ${\dot{q}}_0=h_0(H_{gas}-H_{wall})$ is the honest baseline. Meaning: the ablator is a self-regulating system — no sensors, no valves, no control loop. The physics of blowing automatically caps how much heat can reach the structure, which is precisely why passive ablation is so robust for re-entry and rocket nozzles.

• ${\dot{m}}^{\prime \prime }$ from a heat load and $Q^{\ast }$? ::: ${\dot{m}}^{\prime \prime }={\dot{q}}_{net}/Q^{\ast }$.
• Recession speed from ${\dot{m}}^{\prime \prime }$? ::: $\dot{s}={\dot{m}}^{\prime \prime }/{\rho }_v$.
• Blowing parameter, and what $c_p,h_0$ mean? ::: $B^{\prime }={\dot{m}}^{\prime \prime }{c}_p/h_0$; $c_p$ is the injected gas's specific heat, $h_0$ the un-blown transfer coefficient.
• Blowing reduction factor and its shape? ::: $\ln (1+B^{\prime })/B^{\prime }$; it saturates (diminishing returns).
• Three terms of $Q^{\ast }$? ::: pyrolysis $Q_p$, sensible $c_p(T_w-T_i)$, blowing/blocking $\eta (h_0/{\dot{m}}^{\prime \prime })(H_{gas}-H_w)$.
• Why enthalpy $H$ not temperature $T$ at high $T$? ::: dissociation/recombination adds chemical energy the driving potential must include.
• The self-limiting loop? ::: more heat → more ${\dot{m}}^{\prime \prime }$ → higher $B^{\prime }$ → less $h$ → less heat.