Question bank — Hypergolic propellants — N2O4 - UDMH, MMH

False — they self-ignite into a controlled, sustained burn; an actual explosion (hard start) only happens when the delay is too long and propellant pools first, which is the opposite of what hypergolicity buys you.

False — a short delay means propellant lights the instant it arrives, so almost none accumulates; hard starts come from long delays that let unburnt propellant pool before it flashes over.

False — they are chosen for storability and restart reliability; their $I_{sp}$ (~300–340 s) is well below LOX/LH₂ (~450 s) because the exhaust is heavy (CO₂, N₂), i.e. large $M$.

False — 21 °C is near room temperature, so a modest tank pressure or mild cooling keeps it liquid for years; that is exactly why it counts as storable and cryogens do not.

False — $v_e\propto \sqrt{T_c/M}$ (with $T_c$ the chamber temperature and $M$ the mean exhaust molecular mass), so if burning hotter also produces heavier molecules, the extra $T_c$ can be cancelled or outweighed; that is why the best mixture is often slightly fuel-rich to keep $M$ low.

False — MMH is denser-impulse and cleaner for small RCS thrusters, while UDMH tolerates the higher temperatures of big boosters (Proton, Titan); the mission picks the molecule.

False in principle — the whole point is that contact is the ignition, so there is no spark hardware to fail; this fewer-parts reliability is the design's core advantage.

False — thermodynamics drives nitrogen to the extremely stable N≡N triple bond, so it exits as harmless N₂; forming that bond is the source of most of the released energy.

True — the oxidizer (N₂O₄) is carried onboard, so ignition needs neither air nor a spark; contact alone relights the engine after years in space.

Sign error — the delay grows with $E_a$, so it is ${\tau }_{ig}\propto e^{+E_a/R{T}_0}$; the rate constant $k$ carries the minus sign ($k\propto e^{-E_a/RT}$, the Arrhenius Rate Law), and a small delay comes from small $E_a$ making the exponent near zero.

Ignores molecular mass — since $v_e\propto \sqrt{T_c/M}$, the real target is $T_c/M$; a slightly fuel-rich mix that lowers $M$ (the mean exhaust molecular mass) can beat a hotter mix that raises $M$.

Wrong product — nitrogen must pair into $N_2$; the atom N is high-energy and unstable, and it is the release of energy forming N₂ that makes the reaction so hot.

Confuses amount with ratio — Tsiolkovsky gives $\Delta v=v_e\ln (m_0/m_f)$, where $m_0$ is the full (wet) mass and $m_f$ the burnt-out (dry) mass, so it is the ratio $m_0/m_f$ that matters; extra propellant that also adds tank mass can leave the ratio nearly unchanged.

Reversed — $O/F$ is oxidizer ÷ fuel, so the oxidizer is 1.9× the fuel; here 300 kg fuel needs 570 kg oxidizer, not the other way round.

Wrong meaning — $I_{sp}$ in seconds is $v_e/g_0$, where $g_0$ is standard gravity ($9.81{\text{m/s}}^2$) used purely as a unit-fixing constant; $I_{sp}$ is an efficiency, giving exhaust speed ($v_e=I_{sp}{g}_0$), not how long the engine fires.

Contradiction — here $\rho$ is the pocket's density, $c_p$ its specific heat, and $\dot{q}$ the heat-release rate; adiabatic means no heat leaves, so loss is deliberately set to zero, giving the fastest-possible runaway (an upper bound on how quick ignition can be).

Because ${\tau }_{ig}\propto e^{E_a/R{T}_0}$ (with $R$ the gas constant and $T_0$ the initial mix temperature), and when $E_a\to 0$ the exponential $\to e^0=1$, so nothing multiplies the delay upward; the first collisions already clear the tiny barrier and heat runs away instantly.

A long delay lets fuel and oxidizer keep flowing in and pool unburnt; when it finally lights, all of it reacts at once, spiking pressure far above design (a hard start) and possibly rupturing the chamber.

The N–N bond is weak and the molecule is a strong reducing agent, so the first contact with the strong oxidizer N₂O₄ is a violent redox/acid–base event with almost no barrier — exactly the low-$E_a$ condition for instant ignition (see Arrhenius Rate Law).

Exhaust speed scales as $v_e\propto \sqrt{T_c/M}$ because the nozzle turns heat ($\propto c_p{T}_c$, and $c_p\propto 1/M$) into kinetic energy $\frac{1}{2}{v}_e^2$; for the same released energy, spreading it over heavier molecules ($M$ large) gives each less speed — the energy is there, but carried by slow, massive particles.

N₂O₄ and the hydrazines are liquid near room temperature and lose nothing to boil-off, whereas LOX/LH₂ continuously evaporate; storables also need no igniter, so there is no aged component to fail on command.

The N≡N triple bond is one of the strongest in chemistry, so forming it is thermodynamically overwhelmingly favoured; the drive to reach that low-energy state is what releases the heat that powers the exhaust (see Combustion Thermodynamics).

A fuel-rich mix leaves lighter leftover species and lowers the mean molecular mass $M$; since $v_e\propto \sqrt{T_c/M}$, the drop in $M$ can raise $v_e$ more than the slightly cooler $T_c$ lowers it.

In principle it minimises pooling and hard-start risk, but a truly instantaneous, unlimited runaway means all the mixed propellant reacts essentially at once at the injector — a pressure jump that behaves like a detonation (a supersonic reaction front) rather than a smooth deflagration (subsonic, controlled flame); real designs want a delay short but finite so combustion stays in the deflagration regime and organises smoothly in the chamber.

Higher chamber pressure packs the reactants closer, raising the fuel and oxidizer concentrations $[F],[O]$ (moles per unit volume of each), which speeds the reaction and shortens the delay (the rate $\dot{q}\propto [F][O]$); the mixture ratio $O/F$ has a sweet spot — too fuel-rich or too oxidizer-rich starves one reactant, slowing the runaway and lengthening the delay, so hypergolic engines tune $O/F$ and injector pressure to keep the delay in the safe 1–20 ms band.

No contact means no self-ignition — the reaction needs mixing, so a blocked injector gives no thrust or a dangerous partial fill that can hard-start once flow resumes; unlike Solid Rocket Propellants, the reactants are separate until they meet.

The very first molecular collisions clear the near-zero barrier and release $\Delta H$ (the heat per reaction event); that raises $T$ in a tiny pocket, which speeds the next reactions via $k\propto e^{-E_a/RT}$ — the runaway is self-started, needing no external energy.

The bracket $[1-(p_e/p_c{)}^{(\gamma -1)/\gamma }]\to 0$ (with $\gamma$ the exhaust's heat-capacity ratio and $p_e,p_c$ the exit and chamber pressures), so $v_e\to 0$; with no pressure difference across the nozzle there is no expansion to convert heat into directed motion, hence no thrust.

No — Δv depends on $\ln (m_0/m_f)$ (initial mass over final mass), so doubling the ratio adds only $v_e\ln 2\approx 0.69v_e$; the logarithm punishes you, which is exactly why staging (dropping empty tanks to shrink $m_f$) beats simply building one giant tank.