Hypergolic propellants — N2O4 - UDMH, MMH: Exercises

Constants used throughout:

$g_{0} = 9.81 m/s^{2}$ (standard gravity — the number that turns Specific Impulse in seconds into an exhaust speed).
$R = 8.314 J mol^{- 1} K^{- 1}$ (gas constant — appears whenever temperature and energy trade places).

Level 1 — Recognition

Recall Solution L1.1

Answer: (c) Hypergolic. A specific pair: oxidizer $N_{2} O_{4}$ (NTO), fuel MMH ( $CH_{3} -NH-NH_{2}$ ). Why not the others? Cryogens boil away over months — useless after 8 years of storage. Solids cannot be shut off and restarted on command. Only hypergolics are both storable for years and restart reliably on contact.

Recall Solution L1.2

Answer: $N_{2}$ (nitrogen gas). Forming the nitrogen triple bond $N \equiv N$ is enormously exothermic — it is one of the strongest chemical bonds. The whole reaction is "driven downhill" by locking loose nitrogen atoms into that ultra-stable molecule. (See Combustion Thermodynamics.)

Recall Solution L1.3

Answer: the activation energy $E_{a}$ (the energy hill a collision must climb before it reacts). Small $E_{a}$ → the exponent $E_{a} / R T_{0} \to 0$ → $e^{0} = 1$ → tiny delay. Hypergolics are chemically engineered so $E_{a} \approx 0$ : the reaction has almost no hill to climb, so it fires the instant the liquids touch.

Level 2 — Application

Recall Solution L2.1

Formula: $v_{e} = I_{s p} g_{0}$ . This is the definition of specific impulse — it is exhaust speed measured in "seconds of gravity." $v_{e} = 290 \times 9.81 = 2844.9 m/s \approx 2845 m/s .$

Recall Solution L2.2

Step 1 — oxidizer mass. By definition $O / F = m_{o x} / m_{f u e l}$ , so $m_{o x} = (O / F) \times m_{f u e l} = 1.65 \times 420 = 693 kg .$ Step 2 — total. $m_{p r o p} = m_{f u e l} + m_{o x} = 420 + 693 = 1113 kg .$ Why it matters: the tank must be sized for both liquids; the oxidizer here outweighs the fuel.

Recall Solution L2.3

Step 1: $v_{e} = I_{s p} g_{0} = 315 \times 9.81 = 3090.15 m/s$ . Step 2: final mass $m_{f} = 1500 - 900 = 600 kg$ . Step 3: Tsiolkovsky Rocket Equation: $Δ v = v_{e} ln \frac{m _{0}}{m _{f}} = 3090.15 ln \frac{1500}{600} = 3090.15 \times 0.9163 = 2831.5 m/s .$ Why the log? Each kilogram of exhaust pushes a lighter remaining rocket, so gains compound — the ratio, not the amount, sets $Δ v$ .

Level 3 — Analysis

Recall Solution L3.1

Take the ratio so all the messy prefactors cancel: $\frac{τ _{slug}}{τ _{good}} = \frac{e ^{7.0}}{e ^{2.5}} = e^{7.0 - 2.5} = e^{4.5} = 90.0.$ Interpretation: a 4.5 rise in the exponent multiplies the delay by 90×. In that extra time, unburned propellant pools in the chamber — the seed of a hard start. This is why chemists chase the lowest possible $E_{a}$ (link: Arrhenius Rate Law).

Recall Solution L3.2

$\frac{v _{e}^{hyper}}{v _{e}^{cryo}} = \frac{T _{c}^{hyper} / M ^{hyper}}{T _{c}^{cryo} / M ^{cryo}} = \frac{3000/20}{3500/13} = \frac{150}{269.2} = 0.5572 = 0.7465.$ So hypergolic exhaust is about 75% as fast — matching the real $I_{s p}$ ratio ( $\sim$ 320 s vs $\sim$ 450 s). Which factor dominates? The molar mass. Even though cryogens are only slightly hotter, their exhaust is far lighter ( $M = 13$ vs $20$ ). Light molecules fly out faster for the same thermal energy. See figure below.

Figure — Hypergolic propellants — N2O4 - UDMH, MMH

Recall Solution L3.3

Same $T_{c}$ , so $v_{e} \propto 1/ M$ : $\frac{v _{e, B}}{v _{e, A}} = \frac{M _{A}}{M _{B}} = \frac{21}{19} = 1.1053 = 1.0513.$ Point B wins by about 5.1%. Running slightly fuel-rich makes lighter products (more $H_{2}$ , less $CO_{2}$ ), which raises exhaust speed even at equal temperature. This is why real engines pick an $O / F$ slightly below the value that would maximise $T_{c}$ .

Level 4 — Synthesis

Recall Solution L4.1

Step 1 — exhaust speed. $v_{e} = 320 \times 9.81 = 3139.2 m/s$ . Step 2 — required mass ratio. Invert Tsiolkovsky $Δ v = v_{e} ln (m_{0} / m_{f})$ : $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{2500/3139.2} = e^{0.7964} = 2.2176.$ Step 3 — total mass. Here $m_{f} = m_{dry} = 640 kg$ (all propellant is spent), so $m_{0} = 2.2176 \times 640 = 1419.3 kg, m_{p r o p} = m_{0} - m_{f} = 779.3 kg .$ Step 4 — split by $O / F$ . With $m_{o x} = 1.9 m_{f u e l}$ and $m_{f u e l} + m_{o x} = m_{p r o p}$ : $m_{f u e l} (1 + 1.9) = 779.3 \Rightarrow m_{f u e l} = \frac{779.3}{2.9} = 268.7 kg,$ $m_{o x} = 1.9 \times 268.7 = 510.6 kg .$ Check: $268.7 + 510.6 = 779.3 kg .$ ✓

Recall Solution L4.2

Step 1: $v_{e} = 285 \times 9.81 = 2795.85 m/s$ . Step 2: $m_{0} = 940 + 60 = 1000 kg$ , $m_{f} = 940 kg$ . Step 3: $Δ v = 2795.85 ln (1000/940) = 2795.85 \times 0.06188 = 173.0 m/s .$ Bonus — number of pulses: $60/0.25 = 240$ pulses. Because the propellant is hypergolic, each of those 240 restarts is reliable with no igniter — exactly why RCS uses this pair.

Level 5 — Mastery

Recall Solution L5.1

Write the exponent explicitly: $\frac{E _{a}}{R T _{0}}$ . (a) $E_{a} \to 0$ : exponent $\to 0$ , so $e^{0} = 1$ . The exponential factor disappears entirely and $τ_{i g}$ collapses to just the prefactor — the shortest delay physically possible. This is the design target. (b) $T_{0} \to 0^{+}$ : the denominator $R T_{0} \to 0$ , so $E_{a} / R T_{0} \to + \infty$ , and $e^{+ \infty} \to \infty$ . The delay diverges — the propellant refuses to light. Physically: a cold-soaked thruster (deep-space shadow, no heaters) can fail to ignite or produce a violent hard start once it finally does. This is exactly why hypergolic tanks and lines carry heaters.

Recall Solution L5.2

Differentiate $τ_{i g} = C e^{E_{a} / R T_{0}}$ with respect to $E_{a}$ (treat $T_{0}$ constant): $\frac{d τ _{i g}}{d E _{a}} = C e^{E_{a} / R T_{0}} \cdot \frac{1}{R T _{0}} = τ_{i g} \cdot \frac{1}{R T _{0}} .$ Therefore the fractional sensitivity is $\frac{1}{τ _{i g}} \frac{d τ _{i g}}{d E _{a}} = \frac{1}{R T _{0}} .$ Interpretation: raising $E_{a}$ by $R T_{0}$ joules per mole multiplies the delay by $e$ (a factor of 2.718). At $T_{0} = 300 K$ , $R T_{0} = 8.314 \times 300 = 2494.2 J/mol \approx 2.49 kJ/mol$ . So every extra $\sim$ 2.5 kJ/mol of activation energy nearly triples the ignition delay — a razor-thin margin, which is why hypergolic chemistry is so carefully tuned.

Recall Solution L5.3

Fix $a = 1$ and conserve each element.

Carbon: $2 a = c \Rightarrow c = 2.$
Hydrogen: $8 a = 2 d \Rightarrow d = 4.$
Oxygen: $4 b = 2 c + d = 4 + 4 = 8 \Rightarrow b = 2.$
Nitrogen: $2 a + 2 b = 2 e \Rightarrow 2 + 4 = 2 e \Rightarrow e = 3.$ $C_{2} H_{8} N_{2} + 2 N_{2} O_{4} \to 2 CO_{2} + 4 H_{2} O + 3 N_{2}$ Sanity check — count atoms both sides: C: $2 = 2$ ✓; H: $8 = 8$ ✓; N: $2 + 4 = 6$ , right side $3 \times 2 = 6$ ✓; O: $2 \times 4 = 8$ , right side $2 \times 2 + 4 = 8$ ✓. All nitrogen exits as stable $N_{2}$ , as thermodynamics demands.

Connections

Tsiolkovsky Rocket Equation — the $Δ v = v_{e} ln (m_{0} / m_{f})$ used all through L2–L4.
Specific Impulse — where $v_{e} = I_{s p} g_{0}$ comes from.
Arrhenius Rate Law — the $e^{- E_{a} / R T}$ behind ignition delay.
Cryogenic Propellants — LOX-LH2 — the high- $I_{s p}$ comparison in L3.2.
Reaction Control Systems (RCS) — the restart-budget context of L4.2.
Combustion Thermodynamics — why N₂ carries the energy.
Parent: 3.3.50 Hypergolic propellants — N2O4 - UDMH, MMH (Hinglish)