Exercises — Hybrid engines — advantages, disadvantages

Answer: (c). By definition a classical hybrid keeps the fuel as a solid grain and the oxidiser as a fluid injected down a central port.

• (a) is a liquid engine.
• (b) is a solid motor (fuel and oxidiser baked together).
• (d) is the reverse hybrid — physically possible but not the "classical" one asked for.

The oxidiser (${\dot{m}}_{ox}$). The fuel is a solid wall — it only vaporises when hot oxidiser flows over it. Cut the oxidiser and the wall stops feeding fuel, so total flow $\dot{m}\to 0$ and thrust dies. The flowing reactant is the one holding the control knob.

Tool: the thrust equation $F=\dot{m}{v}_e+(p_e-p_a)A_e$, where (from the toolbox) $p_e$ is the exit pressure, $p_a$ the ambient pressure, and $A_e$ the nozzle exit area. Perfectly expanded means $p_e=p_a$, so $(p_e-p_a)=0$ and the whole second term vanishes. Why this tool: thrust is momentum thrown backward each second — mass per second times speed — plus a correction for when the exit pressure doesn't match the outside air. $F=\dot{m}{v}_e=(3.5)(2300)=\boxed{8050\text{N}}$

Tool: ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}$ (see Regression Rate and Boundary Layer Combustion). Why: fuel peels off a wall of area $A_b$ at speed $\dot{r}$; the swept volume per second is $A_b\dot{r}$ (m³/s); multiply by density to get kg/s. ${\dot{m}}_{fuel}=920\times 0.40\times 1.2\times 10^{-3}=\boxed{0.4416\text{kg/s}}$

Tool: $G_{ox}={\dot{m}}_{ox}/A_{port}$. Why: flux = flow squeezed through the hole's cross-section. $G_{ox}=\frac{2.0}{0.010}=\boxed{200{\text{kg⋅s}}^{-1}{\text{m}}^{-2}}$

Step 1 — old thrust. $F_0=({\dot{m}}_{ox}+{\dot{m}}_{fuel})v_e=(2.0+1.0)(2400)=7200$ N. Why: by the thrust equation, each second the total mass $\dot{m}={\dot{m}}_{ox}+{\dot{m}}_{fuel}$ is hurled out at speed $v_e$, carrying momentum $\dot{m}{v}_e$ backward; by Newton's third law the rocket feels that same momentum forward as thrust. With perfect expansion the pressure term is zero, so $F_0=\dot{m}{v}_e$ — hence we add both flows first, then multiply by $v_e$. Step 2 — how does fuel respond? (build the chain, don't skip links.)

• Fuel flow is never independent — it obeys ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}$. Here ${\rho }_f$ and $A_b$ are fixed, so ${\dot{m}}_{fuel}\propto \dot{r}$.
• The regression rate obeys $\dot{r}=a{G}_{ox}^n$, so $\dot{r}\propto G_{ox}^{0.6}$.
• The flux is $G_{ox}={\dot{m}}_{ox}/A_{port}$; with the port fixed, $G_{ox}\propto {\dot{m}}_{ox}$.
• Chain them: ${\dot{m}}_{fuel}\propto \dot{r}\propto G_{ox}^{0.6}\propto {\dot{m}}_{ox}^{0.6}$. So halving the oxidiser multiplies the fuel by $0.5^{0.6}=0.660$, giving new fuel $=1.0\times 0.660=0.660$ kg/s. What it looks like: in the figure, the amber fuel bar drops less steeply than the amber oxidiser bar — because the exponent $0.6<1$ softens the fall. Step 3 — new thrust. $F=(1.0+0.660)(2400)=3984$ N. $\frac{F_{\text{new}}}{F_{\text{old}}}=\frac{3984}{7200}=0.553\approx \boxed{55\%}$ Halving the oxidiser more than halves thrust-fraction-wise? No — thrust drops to $55\%$, i.e. it nearly halves. Both terms fell, but fuel resisted (exponent $<1$), so the total didn't fall all the way to $50\%$. This coupling is the whole point of throttling a hybrid.

Follow the arrows, one physical link at a time:

1. $A_{port}\uparrow$ (port widens as wall recedes).
2. $G_{ox}={\dot{m}}_{ox}/A_{port}\Rightarrow G_{ox}\downarrow$ (same oxidiser, bigger hole → thinner flux).
3. $\dot{r}=a{G}_{ox}^n\Rightarrow \dot{r}\downarrow$ (weaker flux → slower wall recession).
4. ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}$: even though $A_b$ may grow a bit, the flux fall usually dominates, so ${\dot{m}}_{fuel}\downarrow$.
5. $\text{O/F}={\dot{m}}_{ox}/{\dot{m}}_{fuel}$: numerator constant, denominator falling $\Rightarrow$ O/F rises — the engine drifts oxidiser-rich. This drift away from the optimal O/F slightly lowers $I_{sp}$ — the signature hybrid disadvantage.

Chain it, one earned symbol at a time.

1. $G_{ox}=\frac{{\dot{m}}_{ox}}{A_{port}}=\frac{1.6}{8.0\times 10^{-3}}=200$ kg·s⁻¹·m⁻².
2. $\dot{r}=a{G}_{ox}^n=3.0\times 10^{-5}\times 200^{0.5}=3.0\times 10^{-5}\times 14.142=4.243\times 10^{-4}$ m/s.
3. ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}=920\times 0.50\times 4.243\times 10^{-4}=0.1952$ kg/s.
4. $\text{O/F}=\frac{{\dot{m}}_{ox}}{{\dot{m}}_{fuel}}=\frac{1.6}{0.1952}=8.20$.
5. $\dot{m}={\dot{m}}_{ox}+{\dot{m}}_{fuel}=1.6+0.1952=1.7952$ kg/s.
6. $F=\dot{m}{v}_e=1.7952\times 2350=\boxed{4219\text{N}}$ (about $4.22$ kN). Notice the O/F of $8.2$ — hybrids typically run oxidiser-rich; this is normal.

Why this formula: $I_{sp}$ measures thrust delivered per unit weight-flow of propellant — a fuel-economy score in seconds. $I_{sp}=\frac{F}{\dot{m}{g}_0}=\frac{4219}{1.7952\times 9.81}=\frac{4219}{17.61}=\boxed{239.6\text{s}}$ Since we assumed perfect expansion, $F=\dot{m}{v}_e$, so equivalently $I_{sp}=v_e/g_0=2350/9.81=239.6$ s — a neat cross-check. Real hybrids land in the $250$–$350$ s range; our simplified number sits at the low end, as expected for the model.

Why "just $5\times$ the oxidiser" fails: Thrust $\approx \dot{m}{v}_e$ and $\dot{m}$ is dominated by ${\dot{m}}_{ox}$, so naïvely $5\times$ oxidiser does raise thrust — but at fixed geometry it also slams the O/F far off optimum. Here is the one-sentence derivation of why the O/F climbs: start from $\text{O/F}={\dot{m}}_{ox}/{\dot{m}}_{fuel}$; at fixed port area ${\dot{m}}_{fuel}\propto \dot{r}\propto G_{ox}^n\propto {\dot{m}}_{ox}^n$, so substituting gives $\text{O/F}\propto {\dot{m}}_{ox}/{\dot{m}}_{ox}^n={\dot{m}}_{ox}^{1-n}$, and since $n<1$ the exponent $1-n>0$ means O/F grows as oxidiser grows. Running very oxidiser-rich wastes fuel, drops combustion efficiency, and cooks the nozzle. So thrust is not the only thing that changes — the mixture degrades. The recommended fix (from the parent note): a multi-port grain — carve several parallel ports so total burn area $A_b$ rises, letting ${\dot{m}}_{fuel}$ keep pace with ${\dot{m}}_{ox}$ and hold O/F near optimum. One port simply can't offer enough $A_b$ at high thrust. Burn-area estimate at fixed O/F: at fixed O/F, ${\dot{m}}_{fuel}$ must also scale $\times 5$. Since ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}$, if we hold $\dot{r}$ (by holding flux $G_{ox}$ roughly fixed across the enlarged geometry) then $A_b$ must go $\times 5$: $A_b^{\text{new}}\approx 5\times 0.50=\boxed{2.5{\text{m}}^2}$ That much extra wall area is exactly what a multi-port (wagon-wheel) grain provides.

Take the limit link by link (with $n>0$):

• $G_{ox}={\dot{m}}_{ox}/A_{port}\to 0$.
• $\dot{r}=a{G}_{ox}^n\to a\cdot 0^n=0$ (for $n>0$).
• ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}\to 0$ — the solid stops vaporising because no hot oxidiser bathes it.
• $\dot{m}={\dot{m}}_{ox}+{\dot{m}}_{fuel}\to 0+0=0$.
• $F=\dot{m}{v}_e+(p_e-p_a)A_e\to 0$ (chamber depressurises, so the pressure term collapses too). In words: closing the valve drives every term to zero, smoothly. There is no residual burn, because the fuel has no independent oxidiser (unlike a solid motor). Re-open the valve and oxidiser flux returns → $\dot{r}$ climbs off zero → combustion resumes. That is a clean stop and restart — mathematically forced by $\dot{r}=a{G}_{ox}^n$ vanishing at $G_{ox}=0$.

Which single physical fact generated every advantage and disadvantage on this ladder? ::: Fuel and oxidiser are stored in different phases and physically separated, with the oxidiser valved — everything (safety, throttling, O/F shift, low regression, multi-port scaling) flows from that.