Exercises — Solid propellants — fuel + oxidizer in polymer matrix

• ${\text{NH}}_4{\text{ClO}}_4$ = oxidizer — it supplies the oxygen so the mixture burns without atmosphere.
• Aluminium = fuel/metal — burns fiercely, releases most of the energy.
• HTPB = polymer binder — glues the powder into a solid, castable grain and is itself a hydrocarbon fuel that burns. This is the mnemonic O-F-B.

You change $A_b$, the burning surface area. Since $\dot{m}\propto A_b$ and thrust $F=\dot{m}{v}_e$, the shape of the burning surface over time is what sets the thrust curve. ${\rho }_p$ and $r$ (at fixed pressure) don't depend on port shape.

Convert $r=6\text{mm/s}=0.006\text{m/s}$ (always turn millimetres into metres before multiplying SI quantities). (a) $\dot{m}={\rho }_p{A}_{b}r=1750\times 0.040\times 0.006=0.42\text{kg/s}$. Why: volume eaten per second $=A_{b}r$; multiply by density for mass. (b) Since $p_e=p_a$ the pressure term is zero, so $F=\dot{m}{v}_e=0.42\times 2300=966\text{N}$.

Full formula: $F=\dot{m}{v}_e+(p_e-p_a)A_e$. Momentum part unchanged: $\dot{m}{v}_e=966\text{N}$. Pressure part: $(p_e-p_a)A_e=(20000-0)\times 0.012=240\text{N}$. $F=966+240=1206\text{N}$. Why: in vacuum ($p_a=0$) the exit pressure gives a bonus push — this is why rockets are stronger in space than at sea level.

Take a ratio so the constant $a$ cancels: $\frac{r_2}{r_1}={\left(\frac{p_2}{p_1}\right)}^n=(7/5{)}^{0.35}=1.4^{0.35}.$ $\ln 1.4=0.33647$; $\times 0.35=0.11777$; $e^{0.11777}=1.1250$. $r_2=8\times 1.1250=9.0\text{mm/s}$. Why an exponent, not a simple multiply? Vieille's law is a power law: doubling pressure does not double burn rate. A 40% pressure rise gives only ≈13% more burn rate because $n<1$.

Gas is produced at rate ${\dot{m}}_{\text{gen}}={\rho }_p{A}_{b}a{p}_c^n$ — rises like $p_c^n$. Gas leaves the throat at a rate that (for choked flow) rises like $p_c^1$ — linearly. If pressure blips up by a small $\delta$:

• generation grows by a factor $\propto p_c^n$ → slope $n$,
• discharge grows by a factor $\propto p_c^1$ → slope $1$. If $n<1$ discharge outruns generation → pressure falls back → stable. If $n\ge 1$ generation outruns discharge → pressure climbs further → runaway / detonation. Look at the figure: where the "make gas" curve is below the "dump gas" line past the operating point, the motor self-corrects.

(a) $\dot{m}=1800\times 0.10\times 0.007=1.26\text{kg/s}$; $F=\dot{m}{v}_e=1.26\times 2500=3150\text{N}$. (b) Neutral (constant) thrust, so $I=Ft=3150\times 6.0=18900\text{N⋅s}$. Why $I=Ft$ works here: only because $A_b$ (hence $F$) is constant — that is the whole point of a star port. (c) Propellant burned $=\dot{m}\times t=1.26\times 6.0=7.56\text{kg}$.

$I_{sp}=v_e/g_0=2500/9.81=254.8\text{s}$. (See Specific Impulse.) Check: $I_{sp}{g}_0{m}_{\text{prop}}=254.8\times 9.81\times 7.56=2500\times 7.56=18900\text{N⋅s}$ — matches Q7(b). Why the two definitions agree: $I=Ft=\dot{m}{v}_{e}t=v_e(\dot{m}t)=v_e{m}_{\text{prop}}$, and $v_e=I_{sp}{g}_0$.

Final mass $m_f=40-7.56=32.44\text{kg}$. $\Delta v=2500\ln (40/32.44)=2500\times \ln (1.23305)=2500\times 0.20951=523.8\text{m/s}$. Why a logarithm? Because as fuel burns the vehicle keeps getting lighter, so each later kilogram of exhaust accelerates a smaller ship more — the running total is a log, not a straight line.

(a) $A_b\propto r_{\text{port}}$ (with $L$ fixed), so the factor is $30/20=1.5$. (b) Thrust rises by $1.5-1=0.5=50\%$. Thrust growing during the burn is a progressive curve. (c) A star port adds surface (the points burn back) exactly where the plain tube would lose it, holding $A_b$ ≈ constant → neutral (flat) thrust. See the geometry below.

Burn time from $I=Ft$: $t=30000/5000=6.0\text{s}$. $\dot{m}=F/v_e=5000/2500=2.0\text{kg/s}$. Mass burned $=\dot{m}t=2.0\times 6.0=12.0\text{kg}$. Area from $\dot{m}={\rho }_p{A}_{b}r$: $A_b=\frac{\dot{m}}{{\rho }_{p}r}=\frac{2.0}{1800\times 0.007}=\frac{2.0}{12.6}=0.1587{\text{m}}^2$. Why work backwards? Requirements fix $F$ and $I$; you then solve for geometry — that is exactly how a real solid motor is designed: the grain shape is chosen to hit the target thrust curve.

$\frac{r_2}{r_1}=(6.6/6{)}^{0.40}=1.1^{0.40}$. $\ln 1.1=0.09531$; $\times 0.40=0.038124$; $e^{0.038124}=1.03886$. So $r$ (and $\dot{m}$, and thrust) rise by ≈ $3.9\%$ for a 10% pressure rise. Because $n=0.40<1$, the 10% disturbance produces a smaller 3.9% response — the motor is self-limiting and stable. Good design.

Thrust from burn data ::: $F=\dot{m}{v}_e$, with $\dot{m}={\rho }_p{A}_{b}r$ Add the pressure term when? ::: whenever $p_e\ne p_a$: $F=\dot{m}{v}_e+(p_e-p_a)A_e$ Total impulse of a neutral motor ::: $I=Ft$ (area under a flat thrust curve) Small pressure change ⇒ burn-rate change ::: multiply fractional pressure change by $n$ Why star port over round port ::: keeps $A_b$ constant ⇒ neutral (flat) thrust