Solid propellants — AP - HTPB - Al composition; burn rate dependence on pressure (Vieille's law): Visual walkthrough

This page is the companion to the parent topic. We lean on Heat Conduction (Fourier's Law), Ammonium Perchlorate Decomposition, and later connect to Characteristic Velocity c-star.

Step 1 — The moving surface (what "burn rate" even means)

WHAT. Imagine a slab of solid propellant. Its top face is on fire. The fire eats downward into the solid. The face itself slides down at some speed. That speed is the burn rate, written $r$ .

WHY. Before any formula, we must agree on what we are measuring. $r$ is not how fast flames flicker — it is how fast the solid surface retreats, in millimetres per second. Everything the rocket does (how much gas, how much thrust) is downstream of this one number.

PICTURE. Look at the figure. The blue slab is the unburnt solid. The dashed white line is where the surface was one second ago; the solid white line is where it is now. The red arrow of length $r$ is the distance the face slid in that one second.

Figure — Solid propellants — AP - HTPB - Al composition; burn rate dependence on pressure (Vieille's law)

Step 2 — Turning surface-speed into mass-per-second

WHAT. Every second the surface sweeps through a thin layer. That layer has a volume, and the solid has a density, so the layer has a mass. That mass becomes gas. We count it.

WHY. A rocket cares about kilograms of gas per second (call it $\overset{m}{˙}$ ), not about millimetres. So we translate the geometric speed $r$ into a mass flow.

PICTURE. The figure shows a slab of burning area $A_{b}$ (the whole glowing face). In one second the face moves down by $r$ , carving out a thin box of thickness $r$ and face-area $A_{b}$ . Its volume is $A_{b} \times r$ ; multiply by density $ρ_{p}$ to get mass.

So if we know $r$ , we know everything. The rest of the walkthrough is a hunt for $r$ .

Step 3 — Why the surface moves: the flame feeds heat backwards

WHAT. The solid does not spontaneously vanish. A thin flame hovers a tiny gap above the surface. That flame is hot ( $T_{f}$ ). It sends heat back down into the cold solid, warming the surface until it decomposes into gas. That decomposition is the retreat.

WHY. We need a physical cause for $r$ so we can figure out what controls it. The cause is a heat delivery: no heat back, no retreat. So the pace of retreat is set by how much heat arrives per second per square metre.

PICTURE. Three temperature zones stacked vertically:

deep solid at starting temperature $T_{0}$ (cold),
the surface at $T_{s}$ (warm enough to decompose),
the flame at $T_{f}$ (blazing).

The green arrow shows heat flowing downward from flame to surface across a gap of width $δ$ (the stand-off distance — how high the flame floats).

Step 4 — Bookkeeping the heat: two ways to write the same flux

WHAT. We write the heat flux $q^{''}$ (double-prime means "per unit area", i.e. watts per m²) twice: once as "how much the solid needs", once as "how much the flame supplies". Then we set them equal.

WHY. This is the heart of every derivation in physics: conservation. The heat the flame delivers must equal the heat the retreating solid absorbs — nothing is created or lost at the surface. Two expressions for one quantity → an equation we can solve.

The demand side (what the solid needs). Each kilogram of solid must do two things before it leaves as gas:

Warm up from $T_{0}$ to $T_{s}$ — costs $c (T_{s} - T_{0})$ joules per kg, where $c$ is the specific heat of the solid (joules to warm 1 kg of the solid by 1 K).
Decompose — the surface reaction that turns solid into gas either absorbs or releases heat. We bookkeep this as a heat the arriving solid must still absorb, written $Q$ (units J/kg).

Common mistake The sign of

$Q$ — read the balance as a demand, not a total enthalpy Everything in the bracket $[c (T_{s} - T_{0}) + Q]$ is energy the incoming solid must be given to leave as hot gas.

If the surface decomposition is endothermic (needs heat to break bonds), that heat must also be supplied → $Q > 0$ , so it adds to the demand → the flame has to work harder → burn rate is lower.
If the surface decomposition is exothermic (releases heat), it pays part of the warm-up bill itself → we bookkeep this as $Q < 0$ , which subtracts from the demand → the flame supplies less → burn rate is higher. The trap is thinking "exothermic means we add its energy to the flux". No — an exothermic surface reaction reduces the heat the flame must deliver, hence the minus sign. Sign convention: $Q$ = net heat the flame must still provide per kg, positive when the chemistry costs energy.

The solid arrives at rate $ρ_{p} r$ kg per second per m². So the total heat demanded per unit area is:

$q_{needed}^{''} = kg/s arriving per m^{2} ρ_{p} r joules per kg the flame must supply: warm-up + (net) decomposition cost [c (T_{s} - T_{0}) + Q]$

The supply side (what the flame gives). This is Heat Conduction (Fourier's Law): heat crossing a gap = (conductivity $λ$ ) × (temperature drop across the gap) ÷ (gap width).

$q_{supplied}^{''} = \frac{λ ( T _{f} - T _{s} )}{δ}$

WHY Fourier's law and not something else? Because the gap is a thin, nearly still layer of gas — heat crosses it mainly by conduction (molecules jostling neighbours), and Fourier's law is exactly the rule for conduction across a slab.

PICTURE. The figure overlays both arrows on the gap: a red "demand" arrow (energy sinking into the solid) and a green "supply" arrow (energy conducting down from the flame). They must match.

Setting supply = demand:

$ρ_{p} r [c (T_{s} - T_{0}) + Q] = \frac{λ ( T _{f} - T _{s} )}{δ}$

Step 5 — Solve for $r$ (still has the mystery $δ$ in it)

WHAT. Rearrange the boxed balance to put $r$ alone on the left.

WHY. We wanted $r$ ; here it is, expressed through knowable-ish quantities — except $δ$ , the gap width, is still hiding. That gap is where pressure will sneak in.

$r = call this leading constant K \frac{λ ( T _{f} - T _{s} )}{ρ _{p} [ c ( T _{s} - T _{0} ) + Q ]} \cdot \frac{1}{δ}$

Why the leading constant $K$ is (nearly) constant. Look at each symbol in $K$ :

$λ$ (gas conductivity) — pressure-free (Step 4 kinetic-theory box), varies only slowly with temperature.
$T_{f}$ (flame temperature) — set by the propellant's chemistry (its heat release ÷ its heat capacity); for a fixed recipe it is very nearly fixed.
$T_{s}$ — pinned by decomposition chemistry (Step 3 intuition), essentially constant.
$T_{0}, ρ_{p}, c, Q$ — fixed material properties of the chosen propellant (and $T_{0}$ , the storage temperature, is what makes the coefficient $a$ temperature-dependent later).

So the only runner in the whole expression is $1/ δ$ . We have named the constant lump $K$ ; keep it in mind — it becomes most of the coefficient $a$ .

PICTURE. The figure isolates the punchline: $r$ is inversely proportional to $δ$ . Small gap → big burn rate. A short green bar $δ$ gives a tall red bar $r$ ; a long $δ$ gives a stubby $r$ .

Find how $δ$ depends on pressure, and we are done.

Step 6 — How pressure squeezes the flame closer (the $δ \propto P^{- k}$ link, quantitatively)

WHAT. We show that raising the chamber pressure $P$ makes the flame float lower — the gap $δ$ shrinks as a power law, $δ \propto P^{- k}$ , and we pin down where the exponent $k$ comes from.

WHY — the quantitative argument. The stand-off distance is set by a race between two speeds:

how fast heat/diffusion carries the pre-flame reactants downstream, and
how fast the chemistry consumes them.

A standard premixed-flame result says the flame thickness is $δ \sim \frac{α}{S},$ where $α$ is the gas thermal diffusivity (how fast heat spreads, units m²/s) and $S$ is the flame propagation speed (m/s). Now put in the pressure dependence of each.

Diffusivity vs pressure. In $α = \frac{λ}{ρ _{g} c _{g}}$ , both $λ$ and $c_{g}$ are pressure-free, but the gas density obeys the ideal-gas law $ρ_{g} \propto P$ . Hence $α \propto P^{- 1}$ . (Denser gas spreads heat over a shorter distance.)

Flame speed vs pressure. For a premixed flame the propagation speed follows from balancing diffusion against reaction. Sketch of the standard result: the flame speed obeys $S \sim α \times (reaction rate per unit mass)$ — heat must diffuse and the mixture must react, so the speed is the geometric mean of the two rates. With reaction rate $\propto P^{m}$ and $α \propto P^{- 1}$ , this gives $S \propto P^{- 1} \cdot P^{m} = P^{(m - 1) /2}$ . (The often-quoted shorthand " $S \propto P^{m /2}$ " drops the diffusivity's $P^{- 1}$ ; we keep it explicit here so nothing is asserted without a reason.)

Combine, keeping every factor: $δ \sim \frac{α}{S} \propto \frac{P ^{- 1}}{P ^{(m - 1) /2}} = P^{- (1 + \frac{m - 1}{2})} = P^{- \frac{m + 1}{2}} \equiv P^{- k}, k = \frac{m + 1}{2} .$

So $k$ is not a fudge factor — it is fixed by the overall reaction order $m$ of the flame chemistry. For $m$ between $1$ and $2$ this gives $k$ between $1$ and $1.5$ ; both effects (denser gas, faster chemistry) push the flame down, so $k > 0$ always.

Counter-effects (why the real exponent is smaller than the naïve $k$ ). The clean power law assumes $T_{f}$ and the chemistry are frozen. In reality:

Flame temperature drifts. As $P$ changes, dissociation of hot products shifts $T_{f}$ slightly, nudging both $α$ and $S$ — this partly opposes the squeezing and flattens the exponent.
Convective mixing. Real cross-flow and turbulence thin or thicken the layer in ways pure conduction ignores.
Two-stage AP flame. AP burns in a premixed stage and a diffusion stage; their blend changes with $P$ .

The net of all this is that the measured exponent (which we will call $n$ ) lands around $0.2$ – $0.5$ — smaller than the naïve $k = \frac{m + 1}{2}$ precisely because these counter-effects pull it down.

PICTURE. Two side-by-side scenes: low pressure (flame floats high, big $δ$ , sparse dots) and high pressure (flame hugs the surface, tiny $δ$ , dense dots). The stand-off shrinks visibly.

Step 7 — Substitute and collapse everything into $r = a P^{n}$

WHAT. Put $δ \propto P^{- n}$ (the measured stand-off scaling) into the Step-5 result $r = K / δ$ . Dividing by something proportional to $P^{- n}$ is the same as multiplying by $P^{+ n}$ .

WHY. This is the moment the whole scaffold becomes the famous one-liner. All the thermal constants collected in $K$ get bundled together with the geometric constant of the $δ$ – $P$ law into a single coefficient we call $a$ , and the pressure exponent (after the Step-6 counter-effects) is the measured $n$ .

Do the substitution explicitly. Write the stand-off law with its own constant: $δ = D P^{- n}$ , where $D$ is a fixed length-scale constant (metres·Pa $^{n}$ ) carrying all the diffusivity/flame-speed prefactors. Then

$r = \frac{K}{δ} = \frac{K}{D P ^{- n}} = \frac{K}{D} P^{+ n} .$

Now define the coefficient by bundling the two constants:

$a \equiv \frac{K}{D} = \frac{1}{D} \cdot \frac{λ ( T _{f} - T _{s} )}{ρ _{p} [ c ( T _{s} - T _{0} ) + Q ]}$

and the burn-rate law falls out complete:

$r = a P^{n}$

So $a$ is not a mystery: it is the leading heat-balance constant $K$ divided by the flame-geometry constant $D$ . Every ingredient — gas conductivity $λ$ , the temperature drop $T_{f} - T_{s}$ , the solid density $ρ_{p}$ , its specific heat $c$ , the decomposition cost $Q$ , and the storage temperature $T_{0}$ hidden inside $K$ — lives inside $a$ . That is exactly why $a$ is temperature-dependent (through $T_{0}$ , which sets how much warm-up each kilogram needs) while $n$ is set by the flame chemistry (through $m$ ).

PICTURE. The final derivation figure shows the assembly line: the heat balance box → solve for $r = K / δ$ → the $1/ δ$ throttle → the $δ = D P^{- n}$ substitution → out pops $r = a P^{n}$ with $a = K / D$ . See Vieille's Law for the parent statement.

Step 8 — Validity limits: where the clean law breaks (very low and very high $P$ )

WHAT. $r = a P^{n}$ with a single constant $a$ and single exponent $n$ is an idealisation valid over a middle band of pressures. We name what breaks at each end.

WHY. Every model has a domain. A designer who extrapolates the fitted $a, n$ outside the tested band will mispredict the motor — sometimes dangerously. So we must show the reader the edges.

Low-pressure edge (below the band). As $P$ drops, the stand-off $δ$ grows large, the flame floats far away, and our Step-4 assumption — that conduction across a thin still gap dominates — fails:

Radiative and convective losses from the surface become comparable to the shrinking conductive input, so the surface can't stay at $T_{s}$ .
Below a low-pressure deflagration limit the flame can no longer sustain itself and combustion extinguishes entirely — $r$ does not follow $a P^{n}$ down to zero; it simply stops.

High-pressure edge (above the band). As $P$ climbs very high:

Product dissociation reverses and condensation of species (e.g. metal oxides) changes $T_{f}$ and the effective chemistry, so the constants $a, n$ drift.
The flame structure itself changes (the two AP stages merge; the gas gets so dense the premixed/diffusion balance shifts), so a single $n$ no longer fits — real data often show a "slope break", sometimes even a locally negative- $n$ ("plateau"/"mesa") region engineered on purpose for stability.

PICTURE. A $ln r$ vs $ln P$ plot: a straight central segment (slope $= n$ , the valid band) flanked by a low- $P$ region that peels down to extinction and a high- $P$ region that bends / breaks slope. The straight part is the only region where a single $r = a P^{n}$ is trustworthy.

Step 9 — The feedback edge cases: why $n < 1$ is mandatory

WHAT. We check the extreme values of the exponent, because a formula you don't stress-test can hide a disaster.

WHY. $n$ controls a feedback loop: burning makes gas → gas raises $P$ → higher $P$ raises $r$ → more burning. Whether this loop settles or runs away depends entirely on $n$ .

The two nozzle symbols we are about to use:

The outflow through the throat is $\overset{m}{˙}_{o u t} = \frac{P A _{t}}{c ^{*}}$ : bigger pressure or bigger drain-hole vents more gas; a more efficient propellant (larger $c^{*}$ ) vents less mass for the same push.

Setting generation ( $ρ_{p} A_{b} a P^{n}$ ) equal to outflow gives the equilibrium pressure

$P_{e q} = (\frac{ρ _{p} a c ^{*} A _{b}}{A _{t}})^{\frac{1}{1 - n}} .$

Case A — $n = 0$ (flat, pressure-independent). Then $r = a$ , a constant. The flame gap doesn't care about pressure. Burn rate never speeds up. Perfectly stable, but rarely how real propellants behave.

Case B — $0 < n < 1$ (the good regime, e.g. 0.3–0.5). A pressure bump makes $r$ rise, but sub-proportionally — the curve bends over. The feedback self-corrects: the nozzle vents the extra gas faster than the surface can make it. The exponent $\frac{1}{1 - n}$ is positive and finite.

Case C — $n \geq 1$ (forbidden). The exponent $\frac{1}{1 - n}$ blows up (division by zero at $n = 1$ , negative beyond). Every pressure rise now makes more extra gas than the nozzle can dump → pressure climbs → more gas → runaway → explosion. This is why designers keep $n < 0.5$ .

PICTURE. Three burn-rate-vs-pressure curves on one axis: flat ( $n = 0$ ), gently bending ( $n = 0.35$ , green, safe), and steep ( $n = 1$ , red, runaway). The red one's slope keeps outrunning the nozzle.

The one-picture summary

WHAT'S NEW HERE. This last figure is not a re-run of the earlier scenes — it overlays the causal chain as a single flow and marks where each pressure exponent enters and where each constant lands in $a = K / D$ , so you can trace one arrow all the way from "press harder" to " $r = a P^{n}$ " and read off why the danger threshold sits at $n = 1$ .

One glance: press harder ( $P ↑$ ) → gas denser ( $ρ_{g} \propto P$ ), chemistry faster ( $rate \propto P^{m}$ ) → stand-off $δ = D P^{- n}$ collapses → Fourier heat flux $λ (T_{f} - T_{s}) / δ$ rises → surface heat balance forces $r = K / δ$ up → bundle $a = K / D$ to read $r = a P^{n}$ — and the nozzle's venting law $P A_{t} / c^{*}$ only holds it steady while $n < 1$ .

Recall Feynman retelling — the whole walkthrough in plain words

A block of rocket fuel is on fire on top. The fire isn't touching the block — it hovers a hair's-breadth above it, like a tiny sun. That little sun beams heat down into the block, cooking the top layer until it turns to gas and floats away. So the top of the block sinks, slowly, like an ice cube melting from above. How fast it sinks is the burn rate $r$ .

The heat has to do two jobs on each scoop of solid: warm it up, and pay the chemistry bill for turning solid into gas (that bill is $Q$ — and if the chemistry gives back heat, the bill is negative and helps). Once you know how much heat arrives, you know how fast the block sinks. We call the whole "how fast per unit closeness" bundle $K$ .

How fast it sinks depends on one thing: how close that little sun is. Close sun = fast sinking. Far sun = slow sinking. So the whole question becomes: what sets the distance to the flame? Answer: pressure. Squeeze the gas inside the rocket harder and the flame gets shoved down closer to the surface (denser gas spreads heat over a shorter distance, and the chemistry finishes sooner) — so the block sinks faster. Writing it neatly gives $r = a P^{n}$ : " $a$ " is just $K$ divided by the flame-distance constant $D$ (all the boring properties of your fuel), " $P$ " is how hard you're squeezing, and " $n$ " is how much the flame drops per squeeze after the messy real-world effects flatten it.

Two warnings. First, this neat rule only holds in a middle band of pressures — squeeze too little and the fire goes out; squeeze too much and the chemistry changes and the rule bends. Second, if $n$ is too big (near 1), every squeeze makes the fire so much faster that it squeezes itself harder, faster, harder — and the whole thing detonates. The nozzle (a drain hole of area $A_{t}$ ) can normally bleed off the extra gas fast enough to keep things steady, but only while $n < 1$ . So we pick fuels where the speed-up is gentle. A rocket, tamed.

Recall

What exactly is $P$ in Vieille's law? ::: The total chamber (combustion-gas) pressure pressing on the burning surface, in Pa or MPa — not a partial pressure. What is the burn rate $r$ physically? ::: The speed at which the burning surface retreats into the unburnt solid (mm/s). Which single geometric quantity does $r$ depend on inversely? ::: The flame stand-off distance $δ$ ( $r \propto 1/ δ$ ). What two energy costs appear in the surface heat balance? ::: Warming the solid $c (T_{s} - T_{0})$ and the (net) decomposition cost $Q$ per kg; $Q < 0$ if the surface reaction is exothermic. What is the coefficient $a$ built from? ::: $a = K / D$ , where $K = λ (T_{f} - T_{s}) / {ρ_{p} [c (T_{s} - T_{0}) + Q]}$ is the heat-balance constant and $D$ is the stand-off length constant in $δ = D P^{- n}$ . Where does the exponent come from quantitatively? ::: From $δ \sim α / S$ with $α \propto P^{- 1}$ and $S \propto P^{(m - 1) /2}$ , giving ideal $k = (m + 1) /2$ ; counter-effects flatten it to the measured $n \approx 0.2$ – $0.5$ . Why is $λ$ treated as pressure-independent? ::: Kinetic theory: $λ \propto n_{mol} ℓ$ and $n_{mol} \propto P$ while $ℓ \propto 1/ P$ cancel, leaving $λ$ a function of temperature only. What breaks the law at low and high pressure? ::: Low $P$ : radiative/convective losses and a deflagration limit → extinction. High $P$ : dissociation, condensation and flame-structure changes → slope break; single $a, n$ no longer fit. Why must $n < 1$ ? ::: Equilibrium pressure $\propto (\dots)^{1/ (1 - n)}$ ; at $n \geq 1$ the exponent diverges and the pressure feedback runs away. How do $c$ , $c_{g}$ and $c^{*}$ differ? ::: $c$ = solid specific heat (J·kg⁻¹·K⁻¹); $c_{g}$ = gas specific heat at constant pressure (J·kg⁻¹·K⁻¹); $c^{*}$ = characteristic velocity (m/s), a nozzle/combustion efficiency number.