Burn rate r = a·P^n — Vieille's law: Visual walkthrough

Step 1 — The burning surface is a wall that moves

WHAT: we name the moving-wall speed $r$ . WHY: everything in a solid rocket — thrust, pressure, burn time — is downstream of how fast this wall moves. So it deserves its own symbol first. PICTURE: in the figure the magenta line is the burning surface at time $t$ ; the violet dashed line is where it will be one instant later. The gap between them, divided by the time, is $r$ .

Step 2 — To move the wall, you must cook the next layer

Let us build the energy demand of one second of burning, per unit area of surface.

WHAT: in one second the wall sweeps a distance $r$ , so a slab of thickness $r$ (per unit area) is consumed. Its mass is $ρ_{p} \times r$ — this is the mass flux, mass crossing each square metre per second.

WHY multiply by $c_{p} (T_{s} - T_{0})$ ? Each kilogram must be lifted from $T_{0}$ to $T_{s}$ ; that takes $c_{p} (T_{s} - T_{0})$ joules per kilogram. Multiply mass-per-second by joules-per-kilogram → joules per second, i.e. the power demand:

$\overset{q}{˙}_{need} = mass in a m^{3} ρ_{p} layers/s r J per kg per K c_{p} temperature climb (T_{s} - T_{0})$

PICTURE: the shaded band in the figure is the layer swept in one second; the arrow shows it climbing the temperature ladder from $T_{0}$ up to $T_{s}$ .

Step 3 — The flame delivers that heat by conduction

The rule for how fast heat conducts is Fourier's law: heat flux is proportional to how steep the temperature slope is.

WHAT: the temperature drops from $T_{f}$ (at the flame) to $T_{s}$ (at the surface) across the gap $δ$ . The slope of that drop is $(T_{f} - T_{s}) / δ$ .

WHY divide by $δ$ ? A gradient is a change over a distance. Squeeze the same temperature drop into a smaller gap and the slope steepens, so heat pours down faster. That is the whole engine of the law.

$\overset{q}{˙}_{supply} \approx λ \frac{T _{f} - T _{s} temp drop}{gap it drops across δ}$

PICTURE: the orange gradient triangle shows the temperature falling from $T_{f}$ to $T_{s}$ over the distance $δ$ . The steeper the slope (small $δ$ ), the fatter the downward heat arrow.

Step 4 — Steady burning means supply exactly equals demand

WHAT: set the two expressions equal. WHY: it lets us solve for $r$ — the quantity we care about — in terms of everything else.

$ρ_{p} r c_{p} (T_{s} - T_{0}) = λ \frac{T _{f} - T _{s}}{δ}$

Rearrange (divide both sides by $ρ_{p} c_{p} (T_{s} - T_{0})$ ):

$r = \frac{λ ( T _{f} - T _{s} )}{ρ _{p} c _{p} ( T _{s} - T _{0} ) δ}$

Read the message of this formula: $r$ is inversely proportional to $δ$ . Everything else in it (conductivity, densities, temperatures) is roughly fixed for a given propellant. So the only knob left is the standoff distance $δ$ :

$r \propto \frac{1}{δ}$

PICTURE: two panels — a wide gap (weak, cool, slow $r$ ) versus a hugging flame (small gap, fierce, fast $r$ ).

Step 5 — Pressure pulls the flame closer: $δ \propto P^{- n}$

WHAT: a power $P^{- n}$ (not $- P$ , not $e^{- P}$ ). WHY a power and not something else? Reaction rates in a gas scale with concentration raised to a power (two molecules meeting → concentration $^{2}$ ; concentration itself $\propto P$ ). Anything built from "rate $\propto P^{power}$ " collapses to a power law for the distance too. A power law is the natural shape when the underlying physics is "multiply the pressure and the effect multiplies by a fixed factor."

PICTURE: three flames at $P$ , $2 P$ , $4 P$ — each halving-ish its standoff, drawn to scale.

Step 6 — Substitute and collect constants into $a$

WHAT: put $δ \propto P^{- n}$ into $r \propto 1/ δ$ .

$r \propto \frac{1}{δ} \propto \frac{1}{P ^{- n}} = P^{n}$

WHY collect a constant? Everything that isn't $P$ — the conductivity $λ$ , densities, heat capacities, temperatures, the proportionality factor in $δ \propto P^{- n}$ — never changes as pressure wobbles. Bundle them all into one number and call it $a$ :

$r = a P^{n}$

PICTURE: the collapse — a cloud of physical constants funnelling into the single box $a$ , leaving the clean law behind.

Step 7 — The edge cases (never leave a gap)

$n = 0$ : then $P^{0} = 1$ , so $r = a$ — burn rate is totally deaf to pressure. A flat horizontal line. (No real propellant is exactly this, but it's the "no sensitivity" limit.)
$0 < n < 1$ (the useful zone): $r$ grows with $P$ but slower than $P$ itself — a gently rising curve. This is where the usable propellants live and where the motor is stable (generation $\propto P^{n}$ is outrun by nozzle exhaust $\propto P^{1}$ ).
$n = 1$ : $r = a P$ , a straight line through the origin — the knife-edge where generation and exhaust grow in lockstep. Any nudge is neither damped nor amplified.
$n > 1$ : $r$ shoots up faster than $P$ . A pressure bump makes generation outrun exhaust → runaway → the danger of combustion instability.
$P \to 0$ : $r \to 0$ — no push, no flame reaching back, nothing burns. The formula behaves sensibly at the floor.

WHY show all four $n$ shapes? Because the single number $n$ silently decides whether your motor is a controllable engine or a bomb, and the reader must see the four regimes side by side.

PICTURE: all four curves $r$ vs $P$ on one axis, colour-coded, with the stability verdict labelled on each.

The one-picture summary

The whole derivation is a single chain: heat needed = heat supplied → solve for $r$ → $r \propto 1/ δ$ → $δ \propto P^{- n}$ → $r = a P^{n}$ .

Step 2: demand ρ_{p} r c_{p} (T_{s} - T_{0}) = Step 3: supply λ \frac{T _{f} - T _{s}}{δ} \Rightarrow Step 4 r \propto \frac{1}{δ}, Step 5 δ \propto P^{- n} \Rightarrow Steps 6 r = a P^{n}

Recall Feynman retelling — the whole walkthrough in plain words

Picture a candle that burns downward into itself. How fast the fiery face sinks is the burn rate $r$ (Step 1). To sink one more layer, that cold layer has to be cooked from cold ( $T_{0}$ ) up to melting-into-gas ( $T_{s}$ ) — and cooking costs a fixed amount of heat every second (Step 2). That heat is delivered by the flame floating just above, conducting downward across a tiny gap $δ$ ; the smaller the gap, the steeper the temperature drop and the harder the heat pours in (Step 3). In steady burning the heat needed equals the heat delivered, and when you solve for $r$ you find it's just $1/ δ$ times a pile of constants (Step 4). Now squeeze the whole thing with pressure: more pressure packs the gas tighter, reactions finish sooner, so the flame snuggles closer — $δ$ shrinks like $P^{- n}$ (Step 5). Flip it: $r \propto 1/ δ \propto P^{n}$ , sweep every unchanging constant into one letter $a$ , and out pops $r = a P^{n}$ (Step 6). Finally, the exponent $n$ is the personality of the propellant: below 1 it's a calm, controllable engine; above 1 it's a bomb waiting for a pressure hiccup (Step 7).

Recall

Which single quantity is $r$ inversely proportional to? ::: The flame standoff distance $δ$ . Why does raising $P$ shrink $δ$ ? ::: Denser gas → faster reactions → flame front sits closer to the surface. Where do all the physical constants go in the final law? ::: They collect into the single coefficient $a$ . What makes the $P$ -dependence a power law rather than exponential? ::: Reaction rates scale with concentration (hence $P$ ) raised to a power, so the standoff distance follows a power of $P$ . Verdict for $n > 1$ ? ::: Unstable — generation outruns nozzle exhaust, thermal runaway.