Ablative cooling — charring, blowing: Worked examples

Before we start, every symbol used here (all inherited from the parent) in plain words:

Definition Symbol glossary (nothing is used before this line)

$\overset{q}{˙}$ ::: heat flux — joules of heat crossing one square metre of wall each second. Units $W/m^{2} = J/(m^{2} s)$ .
$Q^{*}$ ::: effective heat of ablation — how many joules you can throw away per kilogram of material you burn off. Big $Q^{*}$ = good ablator. Units $J/kg$ .
$\overset{m}{˙}^{''}$ ::: mass loss rate per unit area — kilograms leaving each square metre each second. Units $kg/(m^{2} s)$ . (The two primes just mean "per area".)
$ρ_{v}$ ::: density of the virgin (unburnt) material. Units $kg/m^{3}$ .
$\overset{s}{˙}$ ::: recession rate — how fast (in metres/second) the surface eats inward.
$s$ ::: total recession depth — how far the surface has eaten inward, in metres.
$t$ ::: elapsed time of the burn or re-entry, in seconds.
$L$ ::: shield/liner thickness — the total depth of material available, in metres.
$h_{0}$ ::: convective heat-transfer coefficient without blowing (see Convective Heat Transfer (Stanton number)).
$B^{'}$ ::: blowing parameter — dimensionless ratio of injected mass to what the flow can carry. $B^{'} = \overset{m}{˙}^{''} c_{p} / h_{0}$ .
$c_{p}$ ::: specific heat — joules to warm 1 kg by 1 K.
$Q_{p}$ ::: enthalpy absorbed per kg pyrolyzed (see Heat of Reaction and Pyrolysis). Positive $Q_{p}$ = the reaction soaks up heat (endothermic); negative would mean it releases heat (exothermic).
$T_{i}$ ::: initial temperature of the cold, unheated material, in kelvin (K).
$T_{w}$ ::: wall (surface) temperature — how hot the exposed ablating face is, in kelvin (K).
$T_{g}$ ::: gas temperature — the temperature of the hot boundary-layer gas, in kelvin (K).
$H_{g}$ ::: enthalpy of the hot gas per kg — total heat content (thermal + chemical) of the gas, in J/kg.
$H_{w}$ ::: enthalpy of the gas at the wall per kg — heat content of the gas right at the surface, in J/kg.

The scenario matrix

Every problem this topic can throw sits in one of the cells below. The two axes are what is being asked (down the rows) and which edge of the input space it probes (the last column). Read the figure as the map, the table as the legend; each example that follows is stamped with its cell.

Figure — Ablative cooling — charring, blowing

#	Cell class	Degenerate / limiting edge it probes
C1	Recession from a heat load (forward)	ordinary mid-range numbers
C2	Blowing factor, small $B^{'}$	limit $B^{'} \to 0$ ⇒ factor $\to 1$
C3	Blowing factor, large $B^{'}$	limit $B^{'} \to \infty$ ⇒ factor $\to 0$ (blow-off)
C4	Zero input	$\overset{q}{˙} = 0$ or $\overset{m}{˙}^{''} = 0$ — nothing happens
C5	Energy-budget split	sum of three independent J/kg sinks
C6	Inverse problem	given survival, solve for required $Q^{*}$ or thickness
C7	Enthalpy vs temperature trap	high- $T$ dissociation — must use $H$ , not $T$
C8	Real-world word problem	re-entry shield, pick numbers yourself
C9	Exam twist	doubling injection — diminishing returns
C10	Sign-flip / exothermic edge	negative reaction enthalpy adds to the load

C1 — Recession from a heat load (the bread-and-butter case)

Worked example C1 · Forward recession

A silica-phenolic liner: $ρ_{v} = 1600 kg/m^{3}$ , $Q^{*} = 8.0 \times 1 0^{6} J/kg$ . It faces $\overset{q}{˙}_{n e t} = 4 MW/m^{2}$ for a $45 s$ burn. How deep does the surface recede?

Forecast: guess before computing — will it be closer to 1 mm, 1 cm, or 10 cm?

Mass loss per area. $\overset{m}{˙}^{''} = \overset{q}{˙} / Q^{*}$ . Why this step? $Q^{*}$ is "joules per kg thrown away", so dividing the flux (joules per area·sec) by it gives kg per area·sec. $\overset{m}{˙}^{''} = \frac{4 \times 1 0 ^{6}}{8.0 \times 1 0 ^{6}} = 0.5 kg/(m^{2} s)$
Convert mass to depth. $\overset{s}{˙} = \overset{m}{˙}^{''} / ρ_{v}$ . Why this step? Losing mass per area from a slab of density $ρ_{v}$ means the surface drops by (mass/area)÷(mass/volume) = length per second. $\overset{s}{˙} = \frac{0.5}{1600} = 3.125 \times 1 0^{- 4} m/s = 0.3125 mm/s$
Total depth over the burn. $s = \overset{s}{˙} \times t = 3.125 \times 1 0^{- 4} \times 45 = 0.0141 m \approx 14.1 mm$ . Why this step? The recession rate $\overset{s}{˙}$ is metres eaten per second; the surface recedes at (near) constant rate through the burn, so multiplying by the total time $t$ integrates the rate into a total depth.

Verify: Units chain: $\frac{W/m ^{2}}{J/kg} = \frac{J/(m ^{2} s)}{J/kg} = kg/(m^{2} s)$ ✔. Then $\frac{kg/(m ^{2} s)}{kg/m ^{3}} = m/s$ ✔. A ~1.4 cm loss means a 2 cm liner survives with margin — the middle guess was right.

C2 — Blowing factor at small $B^{'}$ (limit $\to 1$ )

C3 — Blowing factor at large $B^{'}$ (limit $\to 0$ , blow-off)

Worked example C3 · Strong blowing / blow-off

A raging pyrolysis produces $B^{'} = 10$ . What fraction of the bare-wall heating survives?

Forecast: heavy gas injection — is the surviving fraction closer to 0.5, 0.2, or 0.05?

Apply the model. $\frac{h}{h _{0}} = \frac{ln ( 1 + B ^{'} )}{B ^{'}}$ . Why this step? Same film/Couette blowing law from the parent — it is the only relation that turns a blowing parameter $B^{'}$ into a heating reduction, valid at every $B^{'}$ including this large one. $\frac{h}{h _{0}} = \frac{l n ( 11 )}{10} = \frac{2.3979}{10} = 0.2398$
Interpret. Only about $24%$ of the unblown flux still reaches the wall — the gas cushion has choked most of it. This is the onset of the "blow-off" regime.

Verify: As $B^{'} \to \infty$ , $ln (1 + B^{'})$ grows like $ln B^{'}$ (slow) while the denominator grows like $B^{'}$ (fast), so the ratio $\to 0$ . Our $0.24$ at $B^{'} = 10$ is well below the $0.55$ we got at $B^{'} = 2$ in the parent — monotone decreasing ✔.

C4 — Zero / degenerate input

Worked example C4 · The "nothing happens" cases

Two degenerate scenarios that a careless formula can blow up on: (a) $\overset{q}{˙}_{n e t} = 0$ (engine off, cold). (b) $B^{'} = 0$ (no gas injected at all).

Forecast: what should recession and the blowing factor each equal when the driver is zero?

(a) Zero heat flux. $\overset{m}{˙}^{''} = \overset{q}{˙} / Q^{*} = 0/ Q^{*} = 0$ , so $\overset{s}{˙} = 0$ . Why this step? No heat arriving means no bonds broken and no material lost — physically the surface just sits there.
(b) Zero blowing. Naively $ln (1 + 0) /0 = 0/0$ looks undefined — a trap. Why this step? We must take the limit, not plug in blindly. $lim_{B^{'} \to 0} \frac{l n ( 1 + B ^{'} )}{B ^{'}} = 1$ So $h = h_{0}$ : with no gas injected the wall feels the full bare-wall convection. Sensible, not a divide-by-zero disaster.

Verify: L'Hôpital on $ln (1 + B^{'}) / B^{'}$ : derivative of top $= 1/ (1 + B^{'}) \to 1$ , derivative of bottom $= 1$ , ratio $\to 1$ ✔. And $\overset{m}{˙}^{''} = 0$ trivially gives $\overset{s}{˙} = 0$ ✔.

C5 — Splitting the energy budget

Worked example C5 · Three-sink sum

A carbon-phenolic has $Q_{p} = 2.5 \times 1 0^{6} J/kg$ , $c_{p} = 1400 J/(kg\cdotK)$ , surface at $T_{w} = 3000 K$ from initial $T_{i} = 300 K$ , and a blocking (blowing) term of $5.5 \times 1 0^{6} J/kg$ . Find $Q^{*}$ .

Forecast: which of the three sinks do you expect to be biggest?

Sensible heating sink. $c_{p} (T_{w} - T_{i})$ . Why this step? Before a kilogram can pyrolyze it must first be heated from $T_{i}$ up to the surface temperature $T_{w}$ — that warming costs energy. $c_{p} Δ T = 1400 \times (3000 - 300) = 1400 \times 2700 = 3.78 \times 1 0^{6} J/kg$
Sum the three independent sinks. Why this step? Each channel — chemical decomposition ( $Q_{p}$ ), warming ( $c_{p} Δ T$ ), and boundary-layer blocking — removes joules independently, so they add. $Q^{*} = 2.5 \times 1 0^{6} + 3.78 \times 1 0^{6} + 5.5 \times 1 0^{6} = 11.78 \times 1 0^{6} J/kg$

Verify: All three terms carry units $J/kg$ (for term 1: $J/(kg\cdotK) \times K = J/kg$ ) ✔. The blocking term is largest ( $5.5$ of $11.78$ ), confirming the parent's point that blowing is a main cooling channel, not a side effect.

C6 — Inverse problem: design for survival

Worked example C6 · Solve backwards for required

$Q^{*}$ A shield of thickness $L = 25 mm$ and $ρ_{v} = 1450 kg/m^{3}$ must survive $\overset{q}{˙}_{n e t} = 6 MW/m^{2}$ for a $t = 90 s$ re-entry — while losing at most half its thickness. What minimum $Q^{*}$ does the material need?

Forecast: materials range roughly $Q^{*} \sim 1 0^{6}$ to $1 0^{7}$ J/kg — will the requirement land inside that range?

Allowed recession. Lose at most half of $L = 25$ mm: $s_{ma x} = 12.5 mm = 0.0125 m$ . Why this step? This is the design constraint turned into a length budget.
Allowed recession rate. $\overset{s}{˙} = s_{ma x} / t = 0.0125/90 = 1.389 \times 1 0^{- 4} m/s$ . Why this step? Spread the allowed loss evenly over the burn to get the fastest tolerable rate.
Allowed mass loss rate. $\overset{m}{˙}^{''} = ρ_{v} \overset{s}{˙} = 1450 \times 1.389 \times 1 0^{- 4} = 0.2014 kg/(m^{2} s)$ . Why this step? Convert length budget back to a mass budget via density (inverse of C1 step 2).
Required heat of ablation. $Q^{*} = \overset{q}{˙} / \overset{m}{˙}^{''} = 6 \times 1 0^{6} /0.2014 = 2.98 \times 1 0^{7} J/kg$ . Why this step? Rearranging $\overset{m}{˙}^{''} = \overset{q}{˙} / Q^{*}$ for the unknown $Q^{*}$ .

Verify: $\approx 3.0 \times 1 0^{7} J/kg$ is above carbon-phenolic's typical $\sim 1 \times 1 0^{7}$ — so this harsh case demands a premium ablator or thicker shield. Units: $\frac{W/m ^{2}}{kg/(m ^{2} s)} = J/kg$ ✔. Sanity: plug back — at $Q^{*} = 3.0 \times 1 0^{7}$ , $\overset{m}{˙}^{''} = 6 \times 1 0^{6} /3.0 \times 1 0^{7} = 0.2 kg/(m^{2} s)$ , $\overset{s}{˙} = 0.2/1450 = 1.38 \times 1 0^{- 4}$ m/s, over 90 s $= 12.4$ mm ≈ half ✔.

C7 — Enthalpy vs temperature trap

Worked example C7 · Why

$T$ underestimates at high heat Hot gas at $T_{g} = 6000 K$ , wall at $T_{w} = 3000 K$ , $h_{0} = 400 W/(m^{2} K)$ . Naive temperature-based flux versus the enthalpy-based one, where the true enthalpy difference is $H_{g} - H_{w} = 1.8 \times 1 0^{7} J/kg$ and the mass-transfer coefficient is $h_{0} / c_{p}$ with $c_{p} = 2000 J/(kg\cdotK)$ .

Forecast: at 6000 K air is partly dissociated — will the enthalpy-based flux be smaller or larger than the naive one?

Naive temperature flux. $\overset{q}{˙}_{T} = h_{0} (T_{g} - T_{w}) = 400 \times (6000 - 3000) = 1.2 \times 1 0^{6} W/m^{2}$ . Why this step? This is the tempting textbook form — our baseline to beat.
Enthalpy-based flux. Use $\overset{q}{˙}_{H} = (h_{0} / c_{p}) (H_{g} - H_{w})$ . Why this step? When the gas dissociates and recombines, chemical enthalpy — not just temperature — drives the heating; the correct driving potential is the enthalpy difference $H_{g} - H_{w}$ . $\overset{q}{˙}_{H} = \frac{400}{2000} \times 1.8 \times 1 0^{7} = 0.2 \times 1.8 \times 1 0^{7} = 3.6 \times 1 0^{6} W/m^{2}$
Compare. $\overset{q}{˙}_{H} / \overset{q}{˙}_{T} = 3.6/1.2 = 3.0$ — the enthalpy form predicts three times the load. Why this step? We compute the ratio to expose the size of the error a temperature-only estimate would make; a designer who trusted $\overset{q}{˙}_{T}$ would under-build the shield by a factor of three, so quantifying the gap is the whole point of the trap.

Verify: Units of step 2: $\frac{W/(m ^{2} K)}{J/(kg\cdotK)} \times J/kg = \frac{W/(m ^{2} K) \cdot kg}{J} \times \frac{J}{kg}$ reduces to $W/m^{2}$ ✔. The enthalpy flux is larger, exactly the parent's warning: temperature alone underestimates the heat load. See Re-entry Aerothermodynamics.

C8 — Real-world word problem (re-entry shield)

Worked example C8 · Sizing an Apollo-style shield

A capsule heat shield sees a peak $\overset{q}{˙}_{n e t} = 3 MW/m^{2}$ averaged over a $t = 400 s$ re-entry. The ablator is carbon-phenolic: $ρ_{v} = 1450 kg/m^{3}$ , $Q^{*} = 1.2 \times 1 0^{7} J/kg$ . Engineers want a $50%$ safety margin on recession depth. What shield thickness $L$ should they install?

Forecast: re-entry shields are famously thick — guess: 2 cm, 6 cm, or 20 cm?

Mass loss rate. $\overset{m}{˙}^{''} = \overset{q}{˙} / Q^{*} = 3 \times 1 0^{6} /1.2 \times 1 0^{7} = 0.25 kg/(m^{2} s)$ . Why this step? Same forward chain as C1 — flux divided by heat of ablation.
Recession rate. $\overset{s}{˙} = \overset{m}{˙}^{''} / ρ_{v} = 0.25/1450 = 1.724 \times 1 0^{- 4} m/s$ . Why this step? Divide the mass loss per area by the density to turn kilograms lost into metres of surface eaten, exactly as in C1 step 2.
Total recession. $s = \overset{s}{˙} \times t = 1.724 \times 1 0^{- 4} \times 400 = 0.0690 m = 69.0 mm$ . Why this step? Integrate the (near-constant) rate over the whole re-entry time $t$ .
Add margin. With a $50%$ margin the installed thickness is $L = 1.5 \times s = 1.5 \times 69.0 mm = 103.4 mm \approx 103 mm$ (about $10.3 cm$ ). Why this step? Design practice — carry extra material against uncertainty in the heat load, so multiply the bare-survival depth by $1.5$ .

Verify: ~10 cm matches real Apollo-class shields (they were several cm thick) — the "6–20 cm" bracket was right, not 2 cm. Cross-check with C1's numbers: same physics, longer burn ⇒ much deeper loss ✔. Contrast with Regenerative Cooling and Radiative Cooling, which never consume the wall.

C9 — Exam twist: diminishing returns

Worked example C9 · Doubling the injection

An exam asks: "Blowing at $B^{'} = 2$ gives $h / h_{0} = 0.549$ . If you double the gas injection to $B^{'} = 4$ , does the heating halve again?" Compute and comment.

Forecast: intuition says double the gas → half the heat. Trust it or not?

Factor at $B^{'} = 4$ . $\frac{h}{h _{0}} = \frac{l n ( 5 )}{4} = \frac{1.6094}{4} = 0.4024$ Why this step? Directly evaluate the same $ln (1 + B^{'}) / B^{'}$ model at the doubled parameter.
Compare. Going $B^{'} = 2 \to 4$ moved the factor only from $0.549$ to $0.402$ — a drop of about $0.147$ , not a halving. Why this step? The trap is expecting linearity; the $lo g$ saturates.

Verify: Ratio $0.402/0.549 = 0.73$ , i.e. we kept $73%$ of the previous heating despite doubling mass loss — a poor trade. This is the parent's third "common mistake": more blowing does not help linearly — you just erode twice as fast for a modest gain ✔. Compare $B^{'} = 1$ : $ln 2/1 = 0.693$ — the curve keeps flattening.

C10 — Sign-flip edge: an exothermic reaction adds to the load

Worked example C10 · When the chemistry works against you

Not every surface reaction soaks up heat. The char can react with incoming oxygen ( $C + O_{2}$ ), and that reaction releases heat — an exothermic contribution. Model a case where the pyrolysis is only weakly endothermic, $Q_{p} = + 0.8 \times 1 0^{6} J/kg$ , but surface oxidation dumps $Q_{o x} = - 1.5 \times 1 0^{6} J/kg$ back into the wall (the minus sign meaning "heat added, not removed"). Sensible heating is $c_{p} (T_{w} - T_{i})$ with $c_{p} = 1400 J/(kg\cdotK)$ , $T_{w} = 2600 K$ , $T_{i} = 300 K$ . Ignore blowing for this budget. Is the effective heat of ablation still positive, and what does a negative $Q^{*}$ term physically mean?

Forecast: with a heat-releasing reaction in the mix, will $Q^{*}$ come out bigger or smaller than the pure-endothermic case — and could it ever go negative?

Sensible heating sink (always a genuine sink). $c_{p} Δ T = 1400 \times (2600 - 300) = 1400 \times 2300 = 3.22 \times 1 0^{6} J/kg$ . Why this step? The material still has to be warmed from $T_{i}$ to $T_{w}$ regardless of what the chemistry does — this term never changes sign.
Add all reaction enthalpies with their signs. $Q^{*} = c_{p} Δ T + Q_{p} + Q_{o x}$ . Why this step? $Q^{*}$ is a signed sum of joules-per-kg: endothermic terms are positive (they remove incoming heat), exothermic ones are negative (they inject extra heat), so the honest budget keeps the signs. $Q^{*} = 3.22 \times 1 0^{6} + 0.8 \times 1 0^{6} + (- 1.5 \times 1 0^{6}) = 2.52 \times 1 0^{6} J/kg$
Interpret the surviving sign. $Q^{*} = + 2.52 \times 1 0^{6} J/kg$ is still positive but much smaller than a purely endothermic ablator — the oxidation stole $1.5 \times 1 0^{6} J/kg$ of protection. If the oxidation term ever exceeded the two sinks combined, $Q^{*}$ would go negative, meaning the reaction pumps more heat into the wall than the mass loss carries away — a runaway where ablating faster makes things worse. Why this step? The sign of $Q^{*}$ is the go/no-go flag: positive = protective, negative = self-heating catastrophe.

Verify: Units all $J/kg$ ✔. Sanity: drop the exothermic term ( $Q_{o x} = 0$ ) and you recover $3.22 + 0.8 = 4.02 \times 1 0^{6} J/kg$ , larger than our $2.52 \times 1 0^{6}$ — confirming the exothermic reaction lowered the figure of merit exactly as forecast. This is why Boundary Layer Theory oxygen transport to the char surface is a design worry, not a bonus.

The figure above plots the whole blowing curve $ln (1 + B^{'}) / B^{'}$ with cell C2, the parent's $B^{'} = 2$ , C9's $B^{'} = 4$ , and C3 marked. Read left-to-right: it starts at $1$ (no blowing, C4b), sags gently, then flattens toward $0$ — the geometric picture of "diminishing returns" behind C9 and blow-off C3.

Recall

Recall Cover the answers

In C1, why divide $\overset{q}{˙}$ by $Q^{*}$ first? ::: because $Q^{*}$ is joules-per-kg, so flux÷ $Q^{*}$ gives the mass-loss rate per area.
What is $lim_{B^{'} \to 0} ln (1 + B^{'}) / B^{'}$ and why does it matter? ::: it equals $1$ ; it saves the formula from a fake $0/0$ and means "no blowing ⇒ full bare-wall heating".
Why must you sum the three energy sinks in $Q^{*}$ ? ::: pyrolysis, sensible heating, and blocking each remove joules independently.
Why did the enthalpy flux in C7 beat the temperature flux? ::: dissociation stores chemical energy that temperature alone ignores, so $T$ underestimates the load.
Does doubling $B^{'}$ from 2 to 4 halve the heating? ::: no — the $lo g$ saturates; it only drops from $0.549$ to $0.402$ .
What does a negative $Q^{*}$ term (C10) signify? ::: an exothermic reaction adds heat to the wall; if it outweighs the sinks, $Q^{*}$ goes negative and ablation becomes self-heating runaway.