Combustion of hypergolics — N₂O₄ + UDMH - MMH; ignition delay: Exercises

Before we start, one tiny glossary so nothing is used before it is named:

Level 1 — Recognition

L1.1

State what "hypergolic" means, and name the standard oxidiser and the two standard fuels with their molecular formulas.

Recall Solution

Hypergolic = a fuel/oxidiser pair that ignites spontaneously the instant they touch, with no spark, glow-plug or external igniter.

Oxidiser: $N_{2} O_{4}$ (dinitrogen tetroxide, "NTO").
Fuels: UDMH $= C_{2} H_{8} N_{2}$ and MMH $= CH_{6} N_{2}$ (both hydrazine derivatives).

L1.2

Which of these is kinetics (how fast it starts) and which is thermodynamics (how much energy is released): (a) ignition delay $τ_{i g n}$ , (b) adiabatic flame temperature?

Recall Solution

(a) $τ_{i g n}$ → kinetics. It is a time; it tells you how quickly the reaction gets going, governed by the Arrhenius rate law.
(b) Flame temperature → thermodynamics. It is set by how much energy the reaction ultimately releases (from bond enthalpies), not by how fast. These are independent axes — a fast-starting fuel is not automatically a hot one.

Level 2 — Application

L2.1

Balance the combustion of MMH ( $CH_{6} N_{2}$ ) with $N_{2} O_{4}$ to the products $CO_{2}$ , $H_{2} O$ , $N_{2}$ .

Recall Solution

Carbon and hydrogen products are fixed by the fuel: 1 C → 1 CO₂, and 6 H → 3 H₂O. Write $CH_{6} N_{2} + a N_{2} O_{4} \to CO_{2} + 3 H_{2} O + b N_{2}$ Conserve oxygen (WHY: atoms can't be created): left has $4 a$ O atoms, right has $2 + 3 = 5$ . $4 a = 5 \Rightarrow a = \frac{5}{4} .$ Conserve nitrogen: left $= 2 + 2 a = 2 + \frac{5}{2} = \frac{9}{2}$ atoms $\Rightarrow b = \frac{9}{4}$ . Clear fractions by multiplying every coefficient by 4: $4 CH_{6} N_{2} + 5 N_{2} O_{4} \to 4 CO_{2} + 12 H_{2} O + 9 N_{2}$

L2.2

Using that equation, how much NTO (in kg) is needed to fully burn 3.0 kg of MMH at stoichiometry? Molar masses: $M (MMH) = 46 g/mol$ , $M (N_{2} O_{4}) = 92 g/mol$ .

Recall Solution

Stoichiometric coefficients give a mole ratio; convert to a mass ratio (see Stoichiometry and limiting reagent): $O/F = \frac{5 \times 92}{4 \times 46} = \frac{460}{184} = 2.5.$ So each 1 kg of MMH needs 2.5 kg of NTO. For 3.0 kg of MMH: $3.0 \times 2.5 = 7.5 kg of NTO .$

Level 3 — Analysis

L3.1

Explain, symbol by symbol, why the ignition-delay formula carries $e^{+ E_{a} / R T}$ while the reaction rate constant carries $e^{- E_{a} / R T}$ . Then state which way $τ_{i g n}$ moves as $T$ rises.

Recall Solution

Start from the Arrhenius equation: $k = A e^{- E_{a} / R T}$ . Here $A$ is a constant "attempt frequency", and the exponential is the fraction of collisions energetic enough to clear the $E_{a}$ hill. The time to reach runaway is set by how fast the reaction proceeds, so it is inversely proportional to $k$ : $τ_{i g n} \propto \frac{1}{k} = \frac{1}{A} e^{+ E_{a} / R T} .$ Taking one over $e^{- x}$ gives $e^{+ x}$ — that is where the sign flips. Writing $C = 1/ A$ : $τ_{i g n} = C e^{E_{a} / R T}$ As $T$ rises: the exponent $E_{a} / R T$ shrinks (bigger denominator), so $τ_{i g n}$ decreases — a warm engine ignites faster. A cold engine has a large $τ_{i g n}$ : the dangerous "hard start".

L3.2

From measured data, $τ = 8 ms$ at $T_{1} = 250 K$ and $τ = 2 ms$ at $T_{2} = 300 K$ . Find $E_{a}$ of the pre-ignition chemistry.

Recall Solution

Take logs of $τ = C e^{E_{a} / R T}$ : $ln τ = ln C + \frac{E _{a}}{R} \cdot \frac{1}{T}$ . Write it for both points and subtract — this cancels the unknown $ln C$ (WHY: same fuel pair, same $C$ ): $ln \frac{τ _{1}}{τ _{2}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) .$ Plug numbers. The delay drops by a factor of 4, so $ln (8/2) = ln 4 = 1.386$ , and $\frac{1}{250} - \frac{1}{300} = 0.004000 - 0.003333 = 6.667 \times 1 0^{- 4} K^{- 1} .$ $E_{a} = \frac{R l n 4}{6.667 \times 1 0 ^{- 4}} = \frac{8.314 \times 1.386}{6.667 \times 1 0 ^{- 4}} \approx 1.73 \times 1 0^{4} J/mol \approx 17.3 kJ/mol .$ This is a small activation energy — exactly why hypergolics still ignite in the cold vacuum of space.

Figure — Combustion of hypergolics — N₂O₄ + UDMH - MMH; ignition delay

Level 4 — Synthesis

L4.1

A designer measures $τ_{i g n} = 3.0 ms$ at a chamber inlet temperature of 280 K, and knows $E_{a} = 17.3 kJ/mol$ . Predict $τ_{i g n}$ if the inlet is pre-heated to 320 K. Does this help avoid a hard start?

Recall Solution

Use the two-point form (the $C$ cancels again): $ln \frac{τ _{320}}{τ _{280}} = \frac{E _{a}}{R} (\frac{1}{320} - \frac{1}{280}) .$ Compute the bracket: $\frac{1}{320} - \frac{1}{280} = 0.0031250 - 0.0035714 = - 4.464 \times 1 0^{- 4} K^{- 1}$ (negative — hotter means smaller $τ$ ). $\frac{E _{a}}{R} = \frac{17300}{8.314} = 2081 K .$ $ln \frac{τ _{320}}{τ _{280}} = 2081 \times (- 4.464 \times 1 0^{- 4}) = - 0.9290.$ $τ_{320} = 3.0 \times e^{- 0.9290} = 3.0 \times 0.3950 \approx 1.19 ms .$ Pre-heating cut the delay from 3.0 ms to ≈1.2 ms — yes, a shorter, more reliable delay makes a hard start much less likely, because less unburnt propellant can pool before the flame appears.

L4.2

A satellite thruster runs fuel-rich at O/F = 2.0 (not the stoichiometric 2.5 for MMH). If it carries 1.0 kg MMH, (a) how much NTO does it burn, and (b) how much MMH is left unburnt after all the NTO is consumed?

Recall Solution

(a) O/F = 2.0 means 2.0 kg NTO per kg MMH supplied. With 1.0 kg MMH loaded, the tanks carry $1.0 \times 2.0 = 2.0 kg NTO$ . All of this NTO can burn. (b) The NTO is now the limiting reagent (there is less oxidiser than stoichiometry wants). At the stoichiometric ratio, 2.5 kg NTO consumes 1.0 kg MMH, so 1 kg NTO consumes $1.0/2.5 = 0.40$ kg MMH. Therefore 2.0 kg NTO consumes: $2.0 \times 0.40 = 0.80 kg MMH .$ Unburnt MMH left over: $1.0 - 0.80 = 0.20 kg MMH (excess fuel).$ That leftover fuel-rich gas keeps the chamber cooler and boosts specific impulse — the reason engines choose to run rich.

Level 5 — Mastery

L5.1

Two candidate fuels are tested. Fuel A: $τ = 1.5 ms$ at 300 K, $E_{a} = 12 kJ/mol$ . Fuel B: $τ = 1.5 ms$ at 300 K, $E_{a} = 30 kJ/mol$ . The engine may be started when cold-soaked in deep space at 220 K. Which fuel gives the safer (shorter) delay when cold, and by roughly what factor? Interpret the result physically.

Recall Solution

Both are identical at 300 K, so we compare only how much each grows on cooling to 220 K. Using $ln (τ_{cold} / τ_{300}) = \frac{E _{a}}{R} (\frac{1}{220} - \frac{1}{300})$ . The temperature bracket (same for both): $\frac{1}{220} - \frac{1}{300} = 0.0045455 - 0.0033333 = 1.2121 \times 1 0^{- 3} K^{- 1} .$ Fuel A ( $E_{a} / R = 12000/8.314 = 1443 K$ ): $ln \frac{τ _{A}}{1.5} = 1443 \times 1.2121 \times 1 0^{- 3} = 1.749 \Rightarrow τ_{A} = 1.5 e^{1.749} = 8.6 ms .$ Fuel B ( $E_{a} / R = 30000/8.314 = 3608 K$ ): $ln \frac{τ _{B}}{1.5} = 3608 \times 1.2121 \times 1 0^{- 3} = 4.374 \Rightarrow τ_{B} = 1.5 e^{4.374} = 119 ms .$ Fuel A is safer, with the cold delay roughly $119/8.6 \approx 14 \times$ shorter than Fuel B. Physical reading: a low $E_{a}$ means the reaction barely cares about temperature — it stays quick even when cold-soaked. A high $E_{a}$ reaction collapses when cold, risking a catastrophic hard start on restart. For a space engine that must relight after a cold coast, low activation energy is a survival feature.

Wrap-up recall

Recall One-line takeaways (cover the answers)

Mass O/F for MMH+NTO at stoichiometry? ::: 2.5 (from $5 \times 92/4 \times 46$ ).
Why does $τ$ grow when the engine is cold? ::: $τ \propto e^{+ E_{a} / R T}$ ; smaller $T$ ⇒ bigger exponent ⇒ longer delay.
Low $E_{a}$ fuel vs high $E_{a}$ fuel when cold-soaked — which is safer? ::: Low $E_{a}$ ; its delay barely grows on cooling.
Fuel-rich engine: which reactant is limiting? ::: The oxidiser (NTO) runs out first; excess fuel remains.