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

The exponent $n$. If $n$ is small the burn rate barely notices a pressure change; if $n$ is close to $1$ the burn rate reacts violently. The coefficient $a$ only sets the baseline speed, not the sensitivity.

$P^0=1$ for any pressure, so $r=a\cdot 1=a$ — the burn rate is constant, completely independent of pressure. This is the idealised "pressure-independent" propellant; real grains never quite reach it but low-$n$ ones approach it.

• (i) supplies oxygen → AP (ammonium perchlorate, the oxidiser).
• (ii) holds together + burns → HTPB (the rubbery fuel-binder).
• (iii) raises flame temperature → Al (aluminium, high-energy metal fuel).

$r=a{P}^n=4.0\times 6^{0.30}$ Since $0.30$ is not a whole number, we route through the natural log — this is why $\ln$ appears: it converts a messy power into a multiply we can exponentiate back. $6^{0.30}=e^{0.30\ln 6}=e^{0.30\times 1.7918}=e^{0.5375}=1.7118$ $r=4.0\times 1.7118=6.85\text{ mm/s}$

Convert $r$ to SI first: $6.85\text{ mm/s}=0.00685$ m/s (why: ${\rho }_p$ and $A_b$ are in metres, so $r$ must be too or the units clash). $\dot{m}={\rho }_p{A}_{b}r=1750\times 0.35\times 0.00685=4.20\text{ kg/s}$

Take the ratio — this kills $a$: $\frac{r_2}{r_1}={\left(\frac{2P}{P}\right)}^{0.30}=2^{0.30}=e^{0.30\ln 2}=e^{0.2079}=1.231$ So doubling pressure raises burn rate by only about $23\%$. Gentle — that's the point of a low $n$.

Ratio kills $a$, then logs turn the power law into a straight-line slope — this is exactly how $a$ and $n$ are measured in the lab. $n=\frac{\ln (r_2/r_1)}{\ln (P_2/P_1)}=\frac{\ln (9.1/7.0)}{\ln (8/4)}=\frac{\ln 1.3}{\ln 2}=\frac{0.2624}{0.6931}=0.3785$ So $n\approx 0.38$.

With $n$ known, plug either point back into $r=a{P}^n$ and solve for $a$: $a=\frac{r_1}{P_1^n}=\frac{7.0}{4^{0.3785}}=\frac{7.0}{e^{0.3785\times 1.3863}}=\frac{7.0}{e^{0.5247}}=\frac{7.0}{1.6900}=4.14$ So $a\approx 4.14$ mm/s (at 1 MPa). Sanity check with point 2: $4.14\times 8^{0.3785}=4.14\times 2.195=9.09\approx 9.1$ ✓.

Take logs of $r=a{P}^n$: $\ln r=\ln a+n\ln P$ This is $y=c+mx$ — a straight line. So on log–log axes:

• the slope is $n$ (steepness of the line, red in the figure),
• the intercept where $\ln P=0$ (i.e. $P=1$) is $\ln a$ (green marker). From L3.1/L3.2: slope $=0.38$, intercept $=\ln 4.14=1.42$.

Unit care first. $P_{eq}$ formula mixes $a$ (mm/s per MPa${}^n$), densities and $c^{\ast }$ (SI). Keep the whole thing in one consistent system. We use $a$ in m/s with $P$ in Pa. Converting: $a=4.14$ mm/s at 1 MPa means $r(P)=4.14\times 10^{-3}(P/10^6{)}^{0.38}$ m/s with $P$ in Pa. Writing $r=a^{\prime }{P}^n$ in pure SI: $a^{\prime }=4.14\times 10^{-3}\times (10^6{)}^{-0.38}=4.14\times 10^{-3}\times 10^{-2.28}=4.14\times 10^{-3}\times 5.248\times 10^{-3}=2.173\times 10^{-5}$ Now the balance ${\rho }_p{A}_b{a}^{\prime }{P}^n=P{A}_t/c^{\ast }$ rearranges (see Characteristic Velocity c-star) to: $P_{eq}={\left(\frac{{\rho }_p{a}^{\prime }{c}^{\ast }{A}_b}{A_t}\right)}^{\frac{1}{1-n}}$ Inner bracket: $\frac{1750\times 2.173\times 10^{-5}\times 1500\times 0.6}{0.01}=\frac{1750\times 2.173\times 10^{-5}\times 900}{0.01}$ $=\frac{0.03423}{0.01}\times ...$ Step it cleanly: $1750\times 2.173\times 10^{-5}=0.03803$; $\times 1500=57.05$; $\times 0.6=34.23$; $÷0.01=3423$. Exponent $\frac{1}{1-0.38}=\frac{1}{0.62}=1.6129$. $P_{eq}=3423^{1.6129}=e^{1.6129\times \ln 3423}=e^{1.6129\times 8.138}=e^{13.126}=5.02\times 10^5\text{ Pa}\approx 0.50\text{ MPa}$

From $P_{eq}\propto A_b^{1/(1-n)}$, doubling $A_b$ multiplies $P_{eq}$ by $2^{1/(1-n)}=2^{1.6129}=e^{1.6129\times 0.6931}=e^{1.1179}=3.058$ So pressure rises by a factor $\approx 3.06$ — more than double, because the exponent $1.6129>1$. This is why grain geometry so strongly controls chamber pressure, and why a growing burn area during firing (progressive grain) raises pressure.

The fractional change follows from the ratio, which cancels $a$ and the common base rate: $\frac{r_{new}}{r_{old}}=(1.10{)}^n$

• X: $(1.10{)}^{0.35}=e^{0.35\ln 1.10}=e^{0.35\times 0.09531}=e^{0.03336}=1.0339$ → burn rate up $3.39\%$.
• Y: $(1.10{)}^{0.75}=e^{0.75\times 0.09531}=e^{0.07148}=1.0741$ → burn rate up $7.41\%$. Y reacts more than twice as hard to the same pressure blip. Since more burn rate → more gas → more pressure → still more burn rate, X is far safer: its small $n$ damps the feedback loop. Low $n$ = forgiving motor.

Equilibrium pressure carries the exponent $\frac{1}{1-n}$.

• As $n\to 1^{-}$, the denominator $1-n\to 0^{+}$, so the exponent $\to +\infty$. The pressure the motor "wants" to settle at becomes hypersensitive to the tiniest change in $A_b/A_t$ or $a$ — the equilibrium curve becomes a near-vertical wall (see figure). A microscopic disturbance sends $P_{eq}$ to enormous values.
• At exactly $n=1$: gas generation $\propto P$ and gas outflow $\propto P$ have the same slope in $P$. They no longer cross at a self-correcting point — there is no stable balance. Any excess generation over outflow grows without bound → runaway pressure → the motor chuffs or detonates. This is the rigorous reason designers demand $n<0.5$: it keeps the feedback strongly self-correcting.

Only $a$ changes, so $\frac{P_{eq,\text{cold}}}{P_{eq,\text{warm}}}=(0.85{)}^{\frac{1}{1-0.4}}=0.85^{1.6667}=e^{1.6667\times \ln 0.85}=e^{1.6667\times (-0.16252)}=e^{-0.27086}=0.7627$ So a $15\%$ drop in $a$ pulls equilibrium pressure down to about $76\%$ of nominal — a $24\%$ drop. This is temperature sensitivity: cold rockets run at lower pressure (and thus lower thrust and shifted specific impulse), which is why munitions are qualified across a temperature range. The physics of that $a(T_0)$ dependence traces back to the flame heat balance in Combustion Thermodynamics and Heat Conduction (Fourier's Law).

Recognise the symbols (L1) → plug and convert units (L2) → invert data with logs to get $a,n$ (L3) → chain into equilibrium pressure with careful SI (L4) → reason about sensitivity, stability and the $n\to 1$ wall (L5).