Burn rate r = a·P^n — Vieille's law: Question bank

Before you start, recall the whole law in one breath:

$r$ = speed the solid surface recedes (mm/s), not gas speed.
$P$ = chamber pressure pushing on that surface.
$a$ = burn-rate coefficient (chemistry and starting temperature).
$n$ = pressure exponent, the combustion sensitivity, usually $0.2$ – $0.5$ .

True or false — justify

Vieille's law says the exhaust gas speeds up as pressure rises.

False.

r

is the recession speed of the solid surface (mm/s scale), not the exhaust velocity (km/s scale from the nozzle) — those live in Thrust Equation and Specific Impulse, not here.

n = 0

the propellant's burn rate is completely independent of chamber pressure.

True.

r = a P^{0} = a

, a constant — this is the idealised "plateau" propellant whose burn rate never responds to pressure, which is the safest possible case.

Doubling the pressure always doubles the burn rate.

False. It multiplies

r

2^{n}

; only if

n = 1

does doubling

P

double

r

. For a real

n = 0.4

propellant, doubling

P

raises

r

by just

2^{0.4} \approx 1.3 \times

A propellant with

n = 1.2

is more powerful and therefore better.

False.

n > 1

means gas generation (

\propto P^{1.2}

) outruns nozzle exhaust (

\propto P

), so any pressure bump runs away into an explosion — high

n

is dangerous, not desirable.

On a plot of

r

versus

P

using ordinary (linear) axes you get a straight line whose slope is

n

False.

r = a P^{n}

is a curve on linear axes; it becomes straight only on log–log axes, where slope

= n

and intercept

= lo g a

The coefficient

a

is a fixed property of the chemical formula alone.

False.

a

also carries the initial grain temperature, so the same motor has a different

a

(and burn rate) on a cold winter launch versus a hot summer one — this is Temperature Sensitivity of Propellants.

The condition

n < 1

guarantees the motor is thermally stable.

True (for this feedback loop). Generation

\propto P^{n}

grows slower than exhaust

\propto P^{1}

, so a pressure rise removes gas faster than it adds it — the disturbance self-corrects.

n < 1

, the burn rate decreases when pressure increases.

False.

r

still increases with

P

for any

n > 0

;

n < 1

only means it increases more slowly than linearly, which is what makes the motor stable.

Spot the error

r = a P^{n}

, and since power is force times something,

r

must have units of watts."

The error is treating

r

as power.

r

is a speed (mm/s) — the whole law is calibrated so that

a P^{n}

comes out in speed units;

a

is not dimensionless but absorbs whatever units are needed.

"Because reaction rate

\propto P^{2}

and the flame sits closer at high

P

, the exponent

n

must equal

2

The error is confusing the reaction-rate exponent with the burn-rate exponent. The flame standoff

δ \propto P^{- k}

enters through a heat balance, and

r \propto 1/ δ \propto P^{n}

; the resulting

n

is small (

0.2

–

0.5

), not

2

"A cold-soaked motor and a warm one have the same

a

, so they burn identically."

The error is forgetting temperature lives inside

a

. Colder solid needs more heat to reach ignition temperature, so it recedes slower — same

P

, different

r

"To find

n

from two data points, subtract the burn rates and divide by the pressures."

The error is using a linear slope. Because the law is a power law you must take logs:

n = \frac{ln ( r _{2} / r _{1} )}{ln ( P _{2} / P _{1} )}

, dividing to cancel

a

before taking the log-slope.

"Higher pressure means more gas molecules, so the flame moves farther from the surface."

Backwards. Denser gas makes reactions finish in a shorter distance, so the flame sits closer; a closer, hotter flame drives a steeper temperature gradient and faster burning.

"Steady burning means supply heat and demand heat are unrelated."

The error denies the balance that produces the law. Steady recession requires flame heat supplied to the surface to exactly equal the heat demanded to warm the incoming solid — setting supply = demand is what yields

r

Why questions

Why does the burn rate rise with pressure at all, physically?

Denser gas at higher

P

lets the flame sit closer and hotter, steepening the temperature gradient into the cold solid; by Fourier's Law of Heat Conduction more heat conducts back, so the next layer reaches ignition sooner and the surface recedes faster.

Why do engineers deliberately pick propellants with small

n

(

0.2

–

0.5

A small

n

makes burn rate weakly sensitive to pressure, so accidental pressure bumps barely change gas generation and the motor self-stabilises rather than runs away.

Why does taking logs of

r = a P^{n}

help engineers?

It turns the curve into a straight line

lo g r = lo g a + n lo g P

, so slope and intercept read off directly as

n

and

lo g a

from strand-burner data.

Why is

n = 1

called the knife-edge?

n = 1

generation (

\propto P

) and exhaust (

\propto P

) grow at exactly the same rate, so a pressure disturbance neither grows nor decays — the motor sits on the boundary between stable and explosive.

Why isn't

r

the same thing as the mass flow rate of gas out of the nozzle?

r

measures how fast the burning surface moves inward; mass flow depends on

r

and the total burning surface area and propellant density — the geometry is set by Solid Rocket Motor Grain Geometry.

Why does the flame-standoff distance

δ

appear as a power of pressure?

δ

is set by a race between diffusion and reaction; reaction rate scales as a power of concentration (hence of

P

), so

δ \propto P^{- k}

, and since

r \propto 1/ δ

this hands us the power law

r \propto P^{n}

Edge cases

What happens to the stability argument if

n

is exactly

1

Generation and exhaust track each other perfectly, so pressure is neutrally stable — a disturbance neither self-corrects nor explodes, making the design unacceptably risky in practice.

What does

r = a P^{n}

predict at

P = 0

for

n > 0

r = a \cdot 0^{n} = 0

— with zero chamber pressure there is no flame pushing heat back, so the surface does not recede; burning has effectively stopped.

What does

r = a P^{n}

predict at

P = 1

(in the chosen pressure unit) for any

n

r = a \cdot 1^{n} = a

, so

a

is literally the burn rate at unit pressure — a useful anchor for reading off

a

from a graph.

If a propellant showed

n < 0

, what would that mean physically and would it be usable?

It would burn slower at higher pressure, which is thermodynamically unusual; while trivially "stable," such behaviour signals an ill-behaved combustion regime and isn't a normal design target.

At very high pressures some propellants show

n

climbing toward or past

1

. Why is this dangerous?

Entering the

n \geq 1

regime turns a previously stable motor into a runaway one, which is why the safe operating band must be checked across the whole pressure range, not just at one point — see Combustion Instability and Chamber Pressure and Nozzle Throat.

Two identical grains, one launched at

- 2 0^{\circ} C

, one at

+ 4 0^{\circ} C

— same chamber pressure. Do they burn at the same rate?

No. The warmer grain needs less heat to reach ignition temperature, so its effective

a

is larger and it recedes faster; the cold grain burns slower — a pure temperature effect, not a pressure one.

Recall One-sentence summary to test yourself

The trap-free mental model: $r$ is a surface speed, it always rises with $P$ (for $n > 0$ ) but only as $P^{n}$ , and the single number $n$ decides everything — small $n$ = weak response = stable, $n \geq 1$ = runaway.