The burning rate of a solid propellant grain follows Saint-Robert's law $r = a p_c^n$, where pressure exponent $n = 0.50$. When chamber pressure $p_c = 4.0\text{ MPa}$, the burn rate is measured as $r_1 = 8.0\text{ mm/s}$. If chamber pressure increases to $p_c = 9.0\text{ MPa}$, what is the new burning rate $r_2$ (in mm/s)?

Answer Key:

12.0

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Burning Rate Formulation:** Saint-Robert's (Vieille's) law: $r = a p_c^n$. 2. **Stage 2: Ratio of States:** * $\frac{r_2}{r_1} = \left(\frac{p_{c2}}{p_{c1}}\right)^n$. 3. **Stage 3: Substitution:** * $\frac{r_2}{8.0} = \left(\frac{9.0}{4.0}\right)^{0.50} = \sqrt{2.25} = 1.50$. * $r_2 = 8.0 \times 1.50 = 12.0\text{ mm/s}$. 4. **Stage 4: Physical Stability Context:** Because $n = 0.50 < 1.0$, combustion chamber operation is intrinsically stable. 5. **Stage 5: Official Reference:** George P. Sutton, *Rocket Propulsion Elements*, Chapter 12.

← Prev Question Next Question →