Burn rate r = a·P^n — Vieille's law: Worked examples

This page is a complete drill. We build a small map of every kind of question Vieille's law $r = a P^{n}$ can throw at you, then work one example per box so you never meet a case you haven't seen before.

Everything here uses only the parent law from Vieille's law. If a symbol appears, it was defined there or is rebuilt below.

Recall The three symbols we lean on (rebuilt, no assumptions)

$r$ = linear burn rate: how fast the solid propellant surface eats inward, in $mm/s$ . Picture a candle whose flame chews down into the wax.
$P$ = chamber pressure: how hard the hot gas pushes on that surface, in $MPa$ (mega-pascals).
$a$ = the burn-rate coefficient (a number carrying the chemistry and the grain's starting temperature); $n$ = the pressure exponent (how strongly $r$ reacts to $P$ ).

The law: $r = a P^{n}$ .

The scenario matrix

Vieille's law has no negative angles or quadrants — but it does have distinct regimes of the inputs $P$ and $n$ , plus degenerate and limiting cases. Every exam question lives in exactly one cell below.

Cell	Case class	What makes it special	Example
C1	Forward evaluate ( $P > 1$ )	Plug in, $P^{n}$ grows	Ex 1
C2	Forward evaluate ( $P < 1$ )	$P^{n} < 1$ , burn rate drops below $a$	Ex 2
C3	Degenerate $P = 1$	$P^{n} = 1$ so $r = a$ exactly	Ex 3
C4	Degenerate $n = 0$	pressure-independent burn	Ex 3
C5	Extract $n$ from two points	ratio kills $a$ , use logs	Ex 4
C6	Inverse — find $P$ for a target $r$	undo the power with a root	Ex 5
C7	Stability limit $n \to 1$ and $n > 1$	knife-edge / runaway	Ex 6
C8	Real-world word problem (web thickness / burn time)	connects $r$ to time	Ex 7
C9	Exam twist — temperature shifts $a$	winter vs summer launch	Ex 8

The one tool we keep reaching for: the logarithm

Vieille's law hides the unknown inside an exponent ( $P^{n}$ ). To pull an exponent down where we can solve for it, there is exactly one tool: the logarithm, because by definition $lo g (x^{n}) = n lo g x$ — it turns a power into a multiplication. That's the only reason logs appear anywhere on this page. Whenever the unknown is $n$ or is trapped as $P^{n}$ , reach for $lo g$ ; when the unknown is $P$ itself, undo the power with a root instead. The figure below shows why: on ordinary axes the law bends, but on log–log axes it becomes a straight line whose slope is $n$ .

Figure — Burn rate r = a·P^n — Vieille's law

Worked examples

Worked example Example 3 — Both degenerate inputs ·

Cells C3 & C4 (a) A test rig holds exactly $P = 1 MPa$ . Find $r$ for $a = 5, n = 0.35$ . (b) A hypothetical "plateau propellant" has $n = 0$ . What is $r$ , and why is this special?

Forecast: for (a) will $n$ matter at all? For (b) does pressure matter at all?

(a) $r = 5 \times 1^{0.35}$ . Any number to any power, if the base is $1$ , is $1$ : $1^{0.35} = 1$ . Why this step? This is the definition anchor — it's why we say " $a$ is the burn rate at $P = 1$ ." So $r = 5 mm/s$ regardless of $n$ .
(b) $r = a P^{0}$ . Anything (nonzero) to the power $0$ is $1$ : $P^{0} = 1$ , so $r = a = 5 mm/s$ for every pressure. Why this step? $n = 0$ removes $P$ from the equation entirely — the burn rate is pressure-independent. Real "plateau" propellants approach this over a pressure band to make thrust predictable.

Verify: (a) $r = 5$ . (b) $r = 5$ at $P = 1$ , and also $5$ at $P = 100$ — check: $5 \times 10 0^{0} = 5 \times 1 = 5$ . ✔ The two degenerate cells give the same number here only because $a = 5$ ; the reason differs (base=1 vs exponent=0).

Worked example Example 4 — Extract

$n$ and $a$ from two data points · Cell C5 Strand-burner data: $r_{1} = 4 mm/s$ at $P_{1} = 2 MPa$ ; $r_{2} = 6 mm/s$ at $P_{2} = 5 MPa$ . Find $n$ and $a$ .

Forecast: two unknowns, two data points — will $n$ come out below 1 (stable) or above (dangerous)?

Divide the two law-instances to eliminate $a$ : $\frac{r _{2}}{r _{1}} = \frac{a P _{2}^{n}}{a P _{1}^{n}} = (\frac{P _{2}}{P _{1}})^{n} .$ Why this step? $a$ is unknown and appears in both; dividing cancels it, leaving a single equation in $n$ only.
Put in numbers: $\frac{6}{4} = (\frac{5}{2})^{n}$ , i.e. $1.5 = 2. 5^{n}$ .
Take logs to bring $n$ down: $n = \frac{ln 1.5}{ln 2.5} = \frac{0.4055}{0.9163} = 0.4425$ . Why this step? $n$ is trapped as an exponent — the log is the only tool that frees it (turns $2. 5^{n}$ into $n ln 2.5$ ).
Back out $a$ from either point: $a = \frac{r _{1}}{P _{1}^{n}} = \frac{4}{2 ^{0.4425}} = \frac{4}{1.3589} = 2.943 mm/s$ . Why this step? Once $n$ is known, the law is linear in $a$ — just divide.

Verify: Rebuild $r_{2}$ : $a P_{2}^{n} = 2.943 \times 5^{0.4425} = 2.943 \times 2.039 = 6.00 mm/s$ . ✔ Matches the given $r_{2} = 6$ . And $n = 0.44 < 1$ → stable propellant, answering the forecast.

Worked example Example 5 — Inverse problem: find

$P$ for a target $r$ · Cell C6 A design requires a burn rate of $r = 12 mm/s$ . The propellant has $a = 5 mm/s$ , $n = 0.35$ . What chamber pressure $P$ delivers this?

Forecast: we need $r$ more than double $a$ . Since $n$ is small, will the required $P$ be modest or large?

Start from the law and isolate the power: $12 = 5 P^{0.35} \Rightarrow P^{0.35} = \frac{12}{5} = 2.4$ . Why this step? Now the unknown $P$ is inside a power. Here $P$ is the base, not the exponent — so we undo it with a root, not a log.
Raise both sides to $1/ n = 1/0.35 = 2.857$ : $P = 2. 4^{1/0.35} = 2. 4^{2.857}$ . Why this step? $(P^{0.35})^{1/0.35} = P^{1} = P$ . Applying the reciprocal power is exactly "taking the $n$ -th root" — it cancels the exponent.
Evaluate: $P = e^{2.857 l n 2.4} = e^{2.857 (0.8755)} = e^{2.501} = 12.20 MPa$ .

Verify: Forward-check: $r = 5 \times 12.2 0^{0.35} = 5 \times 2.400 = 12.0 mm/s$ . ✔ The forecast is answered: because $n$ is small, doubling the burn rate demanded a large pressure ( $\sim 12 \times$ ) — weak sensitivity cuts both ways.

Worked example Example 6 — The stability knife-edge

$n \to 1$ and beyond · Cell C7 Three propellants: A has $n = 0.9$ , B has $n = 1.0$ , C has $n = 1.2$ . A random $+ 10%$ pressure spike hits each. For each, does gas generation or gas exhaust win, and what happens?

Forecast: which single value of $n$ is the exact tipping point?

Recall the competition (from the parent note): gas generated $\propto r \propto P^{n}$ ; gas exhausted through the throat $\propto P^{1}$ . Why this step? Stability is a race between these two rates when $P$ rises.
Compare growth after a $+ 10%$ spike, i.e. $P \to 1.1 P$ . Generation multiplies by $1. 1^{n}$ ; exhaust multiplies by $1. 1^{1} = 1.1$ .
- A ( $n = 0.9$ ): generation $\times 1. 1^{0.9} = 1.0896$ ; exhaust $\times 1.1000$ . Exhaust wins ( $1.1000 > 1.0896$ ) → spike is bled off → stable.
- B ( $n = 1.0$ ): both $\times 1.1000$ — dead heat → neutral / knife-edge.
- C ( $n = 1.2$ ): generation $\times 1. 1^{1.2} = 1.1211$ ; exhaust $\times 1.1000$ . Generation wins → more gas → higher $P$ → more generation → runaway explosion. Why this step? Whichever factor is larger dictates whether $P$ returns toward equilibrium or accelerates away.

Verify: $1. 1^{0.9} = 1.0896 < 1.1$ (A stable ✔); $1. 1^{1.0} = 1.1$ (B neutral ✔); $1. 1^{1.2} = 1.1211 > 1.1$ (C unstable ✔). The tipping point is exactly $n = 1$ , answering the forecast.

Worked example Example 7 — Real-world word problem: burn time from web thickness ·

Cell C8 A cylindrical grain burns radially outward. The propellant web (the thickness the flame must eat through) is $w = 30 mm$ . During most of the burn the chamber sits at $P = 7 MPa$ , and the propellant is our Ex-1 material ( $a = 5 mm/s$ , $n = 0.35$ ). Estimate the burn time $t_{b}$ .

Forecast: roughly, will this motor burn for seconds or minutes?

Get the burn rate at operating pressure: from Ex 1, $r = 5 \times 7^{0.35} = 9.88 mm/s$ . Why this step? Time depends on how fast the surface recedes — that's $r$ , the very quantity Vieille's law gives.
Connect distance, rate, time. The surface must recede a distance $w$ at speed $r$ , and speed = distance / time, so $t_{b} = \frac{w}{r} = \frac{30 mm}{9.88 mm/s} .$ Why this step? This is the definition of $r$ as a recession speed — the reason we care about $r$ at all is to predict burn time.
Evaluate: $t_{b} = 3.036 \approx 3.0 s$ .

Verify: Units: $mm \div (mm/s) = s$ . ✔ Answer $\sim 3$ s — a plausible small solid motor burn, answering the forecast (seconds). A thicker web or lower pressure would lengthen it.

Worked example Example 8 — Exam twist: temperature shifts

$a$ · Cell C9 The same motor is launched once on a $- 2 0^{\circ} C$ winter morning and once on a $+ 4 0^{\circ} C$ summer afternoon. Cold-soaking lowers the coefficient to $a_{cold} = 4.6 mm/s$ ; the hot grain raises it to $a_{hot} = 5.4 mm/s$ . With $n = 0.35$ and $P = 7 MPa$ , find both burn rates and the percentage change. (Assume $P$ roughly fixed for a quick estimate.)

Forecast: it's the identical rocket — should $r$ really differ? If so, hotter or colder burns faster?

Cold launch: $r_{cold} = a_{cold} P^{n} = 4.6 \times 7^{0.35} = 4.6 \times 1.976 = 9.09 mm/s$ . Why this step? The chemistry factor $a$ carries the grain's starting temperature — this is the "hidden variable inside $a$ " the parent note warned about.
Hot launch: $r_{hot} = 5.4 \times 1.976 = 10.67 mm/s$ .
Percentage increase from cold to hot: $\frac{10.67 - 9.09}{9.09} \times 100% = \frac{1.58}{9.09} \times 100% = 17.4%$ . Why this step? Since $P^{n}$ is common to both, the ratio $r_{hot} / r_{cold} = a_{hot} / a_{cold}$ — the burn-rate change equals the coefficient change.

Verify: $a_{hot} / a_{cold} = 5.4/4.6 = 1.1739$ , i.e. $+ 17.4%$ , matching step 3. ✔ Answering the forecast: yes, the identical rocket burns $\sim 17%$ faster hot — the physics of temperature sensitivity, which you can dig into in Temperature Sensitivity of Propellants.

How the cells connect out

The heat-balance origin of the law lives in the parent note and rests on Fourier's Law of Heat Conduction.
Cell C7 (runaway) is the doorway to Combustion Instability.
Cell C8's burn time feeds Solid Rocket Motor Grain Geometry, Chamber Pressure and Nozzle Throat, and ultimately the Thrust Equation and Specific Impulse.

Recall Quick self-test (cover the answers)

Which cell is " $P < 1$ so $r < a$ "? ::: C2 — a positive power of a base below 1 gives a factor below 1. To find $n$ from two points, what cancels $a$ ? ::: Dividing the two law-instances: $r_{2} / r_{1} = (P_{2} / P_{1})^{n}$ . To solve for $P$ given $r$ , do you use a log or a root? ::: A root (reciprocal power $1/ n$ ) — $P$ is the base, not the exponent. The exact stability tipping point? ::: $n = 1$ : generation $\propto P^{n}$ ties exhaust $\propto P^{1}$ . Why does the same rocket burn faster when hot? ::: A higher grain temperature raises the coefficient $a$ ; $r$ scales directly with $a$ .