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

What: we plug in $P=1$. Why: anything raised to a power, when the base is $1$, is $1$: $1^{0.4}=1$. $r=a\cdot 1^{0.4}=a=3\text{mm/s}.$ That's the whole point of defining $a$ "at $1\text{MPa}$": $a$ is the burn rate at unit pressure. The exponent never gets a chance to act because $1$ to any power stays $1$.

(i) the exponent $n=0.35$ — it is the sensitivity; a bigger $n$ means pressure changes swing the rate harder. (ii) the coefficient $a$ — the rate when $P=1$. (iii) $r$ itself — the speed the burning surface eats into the solid, not the exhaust speed.

What: direct substitution. Why the exponential appears: $8^{0.3}$ is easiest through $8^{0.3}=e^{0.3\ln 8}$ — the log turns the power into a multiply, then $e$ undoes the log. $8^{0.3}=e^{0.3\times 2.0794}=e^{0.6238}=\mathrm{1.866.}$ $r=5\times 1.866=9.33\text{mm/s}.$ Sense check: pressure went up $8\times$ but $r$ barely doubled — that's exactly what a small $n$ should do.

What: solve $12=6P^{0.5}$ for $P$. Why: divide first to isolate the power, then undo the $0.5$ power by raising to $1/0.5=2$. $P^{0.5}=\frac{12}{6}=2\Rightarrow P=2^{1/0.5}=2^2=4\text{MPa}.$ Because $n=0.5$ means "square root of $P$", to get twice the burn rate you need four times the pressure.

Why the ratio trick: we don't know $a$, but dividing the law at two pressures cancels it completely: $\frac{r_2}{r_1}={\left(\frac{P_2}{P_1}\right)}^n={\left(\frac{9}{3}\right)}^{0.45}=3^{0.45}.$ $3^{0.45}=e^{0.45\times 1.0986}=e^{0.4944}=\mathrm{1.6396.}$ $r_2=7\times 1.6396=11.48\text{mm/s}.$

Step 1 (why divide): the ratio kills $a$ so we can isolate $n$: $\frac{6}{4}={\left(\frac{5}{2}\right)}^n\Rightarrow 1.5=2.5^n.$ Step 2 (why logs): logs pull $n$ down from the exponent: $n=\frac{\ln 1.5}{\ln 2.5}=\frac{0.4055}{0.9163}=\mathrm{0.4426.}$ Step 3 — recover $a$ by putting $n$ back into one point: $a=\frac{r_1}{P_1^n}=\frac{4}{2^{0.4426}}=\frac{4}{1.359}=2.943\text{mm/s}.$ Sense check: $n<1$ → this propellant is stable. ✔

Why a straight line at all: taking ${\log }_{10}$ of $r=a{P}^n$ gives $\log r=\log a+n\log P$ — the equation of a line with slope $n$ and intercept $\log a$. Slope (=$n$): rise over run between the two marked points: $n=\frac{\log 12.6-\log 5}{\log 10-\log 1}=\frac{1.1004-0.6990}{1-0}=0.4014\approx \mathrm{0.40.}$ Intercept (=$\log a$): at $\log P=0$ (i.e. $P=1$), $\log r=\log 5$, so $a=5\text{mm/s}$. This is exactly the extraction method engineers use on real strand-burner runs.

Why compare exponents: at equilibrium generation = exhaust. Bump $P$ up a little; whichever term grows faster wins.

• A ($n=0.85<1$): exhaust ($P^1$) grows faster than generation ($P^{0.85}$). Extra gas leaves faster than it is made → pressure falls back → stable, survives. ✅
• B ($n=1.15>1$): generation ($P^{1.15}$) outpaces exhaust ($P^1$). The bump feeds itself → runaway → explodes. ❌ The single dividing line is $n=1$. This is the 80/20 fact of the whole topic.

Why divide distance by rate: $r$ is the recession speed, so time = distance ÷ speed, exactly like walking distance over walking speed: $t_b=\frac{w}{r}=\frac{30\text{mm}}{9\text{mm/s}}=3.33\text{s}.$ Link: the web $w$ comes from the Solid Rocket Motor Grain Geometry; the value of $r$ comes from Vieille's law given the operating $P$. This is where the burn-rate law meets the actual hardware.

Why the ratio: percent change only needs the factor, and the factor drops $a$: $\frac{r_2}{r_1}={\left(\frac{6}{4}\right)}^{0.4}=1.5^{0.4}=e^{0.4\times 0.4055}=e^{0.1622}=\mathrm{1.176.}$ So $r$ rises by $1.176-1=0.176=17.6\%$. Sense check: pressure rose $50\%$ but rate only $\sim 18\%$ — small $n$ softens the response, as it should.

Why $a$ moves with temperature: a warmer grain starts closer to its ignition temperature, so less heat is needed to burn the next layer → faster recession → larger $a$. This is Temperature Sensitivity of Propellants in action. $5^{0.35}=e^{0.35\times 1.6094}=e^{0.5633}=\mathrm{1.7565.}$ $r_{\text{cold}}=4.6\times 1.7565=8.080\text{mm/s},r_{\text{hot}}=5.4\times 1.7565=9.485\text{mm/s}.$ Since $P$ and $n$ are shared, the ratio is just $a_{\text{hot}}/a_{\text{cold}}$: $\frac{r_{\text{hot}}}{r_{\text{cold}}}=\frac{5.4}{4.6}=1.1739\Rightarrow +17.4\%.$ Consequence: the same rocket burns $\sim 17\%$ faster on a hot day — higher peak $P$, shorter burn. This is why motors are temperature-conditioned before firing.

Stability first: both have $n<1$, so both are stable — that filter alone doesn't decide it. Doubling test: when $P\to 2P$, rate factor $=2^n$: $2^{0.35}=1.274(+27\%),2^{0.95}=1.932(+93\%).$ Y ($2^{0.95}=1.93\approx 2$) nearly doubles — it meets spec (ii). X gives only $+27\%$. The tension: Y is closer to the $n=1$ knife-edge, so it is stable but less forgiving; a real designer weighs "responsiveness" (large $n$) against "safety margin" (distance below $n=1$). Spec-matching alone points to Y here.

Why check the ratios: if one power law fits, then each doubling of $\log P$ must give the same slope. Use the two ratios: $n_1=\frac{\ln (4.9/3)}{\ln (4/1)}=\frac{\ln 1.6333}{\ln 4}=\frac{0.4907}{1.3863}=0.3540,$ $n_2=\frac{\ln (8.0/4.9)}{\ln (16/4)}=\frac{\ln 1.6327}{\ln 4}=\frac{0.4903}{1.3863}=\mathrm{0.3537.}$ The two slopes agree ($\approx 0.354$) → one law fits. From the first point $P=1$: $a=r=3\text{mm/s}$. $\boxed{r=3P^{0.354}}.$ Verify the middle point: $3\times 4^{0.354}=3\times 1.633=4.90$ ✔.

Every problem here is one law, $r=a{P}^n$, read in three directions: forward (plug in $P$), backward (solve for $P$), and sideways (ratios and logs to get $n,a$ without knowing $a$). Stability is the single inequality $n<1$; temperature lives inside $a$; the web-over-rate gives burn time. Master those four moves and D4 is done.