Hypergolic propellants — N2O4 - UDMH, MMH: Worked examples

Everything here rests on three imported facts. Let me restate them so no symbol is unearned.

Formula The three tools (restated so nothing is assumed)

(1) Exhaust speed from specific impulse — see Specific Impulse: $v_{e} = I_{s p} g_{0}, g_{0} = 9.81 m/s^{2}$ Here $I_{s p}$ ("specific impulse", in seconds) is just a repackaging of exhaust speed; multiplying by the constant $g_{0}$ converts it into metres per second. $v_{e}$ is how fast the burnt gas leaves the nozzle.

(2) Tsiolkovsky — see Tsiolkovsky Rocket Equation: $Δ v = v_{e} ln \frac{m _{0}}{m _{f}}$ $m_{0}$ = mass before the burn (dry rocket + all propellant), $m_{f}$ = mass after (dry rocket only, plus leftover). The symbol $ln$ is the natural logarithm — the question " $e$ to what power gives this number?". It appears because velocity builds up in equal fractional steps of mass thrown away, and adding up equal fractional steps is exactly what a logarithm sums.

(3) Mixture ratio bookkeeping: $O / F = \frac{m _{o x}}{m _{f u e l}}, m_{p r o p} = m_{o x} + m_{f u e l}$ $O / F$ ("oh over eff") is simply how many kilograms of oxidizer per kilogram of fuel.

The scenario matrix

Every problem this topic can pose falls into one of these cells. The worked examples below are each tagged with the cell they cover.

#	Cell (case class)	What makes it tricky	Covered by
A	Ordinary Δv — moderate mass ratio	plug-and-chug baseline	Ex 1
B	Large mass ratio limit ( $m_{0} / m_{f} ≫ 1$ )	$ln$ grows slowly — diminishing returns	Ex 2
C	Small mass ratio limit ( $m_{0} / m_{f} \to 1$ )	$ln (1 + x) \approx x$ — tiny burns	Ex 3
D	Degenerate / zero input	$Δ v = 0$ when nothing burns	Ex 3 (part b)
E	Mixture-ratio bookkeeping — solve for tank split	split total mass by $O / F$	Ex 4
F	Ignition-delay ratio (the exponential)	ratio kills the prefactor	Ex 5
G	Real-world word problem (RCS pulse train)	many tiny burns, Reaction Control Systems (RCS)	Ex 6
H	Exam twist — "hotter but heavier" $v_{e} \propto T_{c} / M$	temperature up but $I_{s p}$ down	Ex 7
I	Cross-compare with cryogenic	same Δv, different $I_{s p}$	Ex 8

Figure — Hypergolic propellants — N2O4 - UDMH, MMH

The figure above is the map of this page: the flat middle of the $ln$ curve is where ordinary rockets live (cell A), the far right is the "diminishing returns" tail (cell B), and the region hugging $m_{0} / m_{f} = 1$ is the tiny-burn limit (cells C, D). Keep glancing back at it.

Ex 1 — Cell A · the ordinary case

Worked example A moderate kick burn

A probe stage: $m_{0} = 1200 kg$ , it burns $700 kg$ of N₂O₄/MMH, $I_{s p} = 320 s$ . Find $Δ v$ .

Forecast: Guess before computing — a mass ratio of $1200/500 = 2.4$ . Do you expect $Δ v$ bigger or smaller than $v_{e}$ ? (Since $ln 2.4 \approx 0.88 < 1$ , expect smaller than $v_{e}$ .)

Step 1. $v_{e} = I_{s p} g_{0} = 320 \times 9.81 = 3139 m/s$ . Why this step? Tool (1): $I_{s p}$ in seconds is useless in Tsiolkovsky until converted to m/s.

Step 2. $m_{f} = m_{0} - m_{p r o p} = 1200 - 700 = 500 kg$ . Why this step? Tsiolkovsky needs the final mass, and burnt propellant is gone.

Step 3. $Δ v = v_{e} ln \frac{m _{0}}{m _{f}} = 3139 ln \frac{1200}{500} = 3139 \times 0.8755 = 2748 m/s$ . Why this step? Tool (2): fractional mass thrown, summed by $ln$ .

Verify: Units: $m/s \times (dimensionless log) = m/s$ ✓. Sanity: $2748 < 3139 = v_{e}$ exactly as forecast, because $ln 2.4 < 1$ . ✓

Ex 2 — Cell B · the large-ratio limit

Worked example Why you can't cheat with more propellant

Same stage, same $I_{s p} = 320 s$ , but now the dry mass is only $m_{f} = 120 kg$ and $m_{0} = 1200 kg$ (mass ratio $= 10$ ). Then someone doubles the tanks: $m_{0} = 2400 kg$ , still $m_{f} = 120$ (ratio $= 20$ ). Compare the two $Δ v$ .

Forecast: You doubled the propellant. Did $Δ v$ double? Guess.

Step 1. Ratio 10: $Δ v = 3139 ln 10 = 3139 \times 2.3026 = 7228 m/s$ . Why this step? Baseline for comparison.

Step 2. Ratio 20: $Δ v = 3139 ln 20 = 3139 \times 2.9957 = 9403 m/s$ . Why this step? Tool (2) again with the new ratio.

Step 3. Gain from doubling: $9403 - 7228 = 2175 m/s$ — only a $30%$ increase for a $100%$ propellant increase. Why this step? This is the "diminishing returns" lesson: $ln$ grows like $ln (2 x) = ln x + ln 2$ , so doubling adds a fixed $3139 ln 2 \approx 2176 m/s$ no matter where you start.

Verify: Predicted fixed add-on $= 3139 ln 2 = 3139 \times 0.6931 = 2176 m/s$ , matching the $2175$ above (rounding). ✓ Look at the far-right flattening in the figure — that flat slope is exactly this effect.

Ex 3 — Cells C & D · tiny burns and the zero case

Worked example The near-1 limit and the degenerate zero

(a) An RCS thruster nudges a $500 kg$ spacecraft, expelling just $2 kg$ of N₂O₄/MMH, $I_{s p} = 290 s$ . Find $Δ v$ . (b) What is $Δ v$ if the thruster fails to fire and burns $0 kg$ ?

Forecast: For (a) the mass ratio is $500/498 = 1.004$ — barely above 1. Guess: is exact $ln$ or the shortcut $ln (1 + x) \approx x$ good here?

Step 1 (a). $v_{e} = 290 \times 9.81 = 2844.9 m/s$ . Why this step? Tool (1).

Step 2 (a). $m_{f} = 500 - 2 = 498 kg$ , ratio $= 500/498 = 1.004016$ . Why this step? Set up the log argument.

Step 3 (a). $Δ v = 2844.9 ln (1.004016) = 2844.9 \times 0.0040080 = 11.40 m/s$ . Why this step? Tool (2). Note $ln (1 + x) \approx x$ with $x = m_{p r o p} / m_{f} = 2/498 = 0.004016$ gives $2844.9 \times 0.004016 = 11.43$ — within $0.3%$ . For tiny burns you may skip the log entirely.

Step 4 (b). Burn $0$ kg $\Rightarrow m_{f} = m_{0} = 500 \Rightarrow ln (500/500) = ln 1 = 0 \Rightarrow Δ v = 0$ . Why this step? The degenerate case: $ln 1 = 0$ because $e^{0} = 1$ . No mass thrown, no push — the equation gives exactly zero, not an error. This is the left edge of the figure where the curve touches the axis.

Verify: (a) units m/s ✓; the approximation $x = 0.004016$ matches the exact log to 3 significant figures ✓. (b) $ln 1 = 0$ is exact ✓.

Ex 4 — Cell E · sizing the two tanks

Ex 5 — Cell F · ignition-delay ratio

Worked example Good hypergolic vs sluggish pair

The parent note's delay law is $τ_{i g} \propto e^{E_{a} / R T_{0}}$ (see Arrhenius Rate Law). A good pair has $E_{a} / R T_{0} = 3$ ; a sluggish pair has $E_{a} / R T_{0} = 8$ at the same $T_{0}, ρ, Δ H$ . How many times longer is the sluggish delay?

Forecast: The difference in exponent is $8 - 3 = 5$ . Guess how big $e^{5}$ is — bigger or smaller than 100?

Step 1. The prefactor (the $ρ c_{p} R T_{0}^{2} /Δ H A [F] [O] E_{a}$ lump) is the same for both, so it cancels in the ratio. Why this step? When two things differ only in the exponent, take the ratio so the messy prefactor disappears.

Step 2. $\frac{τ _{s l o w}}{τ _{g oo d}} = \frac{e ^{8}}{e ^{3}} = e^{8 - 3} = e^{5}$ . Why this step? $e^{a} / e^{b} = e^{a - b}$ — the defining property of the exponential.

Step 3. $e^{5} = 148.4 \approx 150 \times$ . Why this step? Evaluate.

Verify: Cross-check with the parent's numbers: $e^{3} = 20.09$ , $e^{8} = 2981$ , and $2981/20.09 = 148.4$ ✓ — the ratio route and the direct route agree. A $150 \times$ longer delay lets propellant pool → hard start risk. ✓

Ex 6 — Cell G · real-world RCS pulse train

Worked example Adding up many small nudges

A Reaction Control Systems (RCS) thruster fires 40 short pulses during a docking, each expelling $0.5 kg$ of N₂O₄/MMH from a $600 kg$ spacecraft, $I_{s p} = 280 s$ . Find the total $Δ v$ .

Forecast: Total propellant $= 40 \times 0.5 = 20 kg$ . Would doing it in one big burn vs 40 small ones change the answer? (Trick: no — Tsiolkovsky only cares about start and end mass.)

Step 1. $v_{e} = 280 \times 9.81 = 2746.8 m/s$ . Why this step? Tool (1).

Step 2. Total propellant $= 40 \times 0.5 = 20 kg$ , so $m_{f} = 600 - 20 = 580 kg$ . Why this step? The individual pulses telescope: $ln \frac{m _{0}}{m _{1}} + ln \frac{m _{1}}{m _{2}} + \dots = ln \frac{m _{0}}{m _{f}}$ . Only endpoints survive.

Step 3. $Δ v = 2746.8 ln \frac{600}{580} = 2746.8 \times 0.033902 = 93.1 m/s$ . Why this step? Tool (2) on the endpoints.

Verify: Compare to one $20$ -kg burn: identical setup, identical $93.1 m/s$ ✓ — confirming pulse count is irrelevant to Δv (it matters for precision, not total impulse). Units m/s ✓.

Ex 7 — Cell H · the exam twist ("hotter but heavier")

Worked example Chamber temperature up, but I_sp down

A student proposes running an N₂O₄/MMH engine more oxidizer-rich to raise chamber temperature from $T_{c} = 3000 K$ to $3300 K$ , but this raises the mean exhaust molecular mass from $M = 20 g/mol$ to $M = 25 g/mol$ . Does $v_{e}$ (and thus $I_{s p}$ ) go up or down? Use $v_{e} \propto T_{c} / M$ (see Combustion Thermodynamics).

Forecast: Hotter feels faster. Guess the direction before computing.

Step 1. Form the ratio $\frac{v _{e, new}}{v _{e, old}} = \frac{T _{c, new} / M _{new}}{T _{c, old} / M _{old}}$ . Why this step? Only the combination $T_{c} / M$ matters; ratio cancels every unknown constant.

Step 2. $T _{c} / M$ : old $= 3000/20 = 150$ ; new $= 3300/25 = 132$ . Why this step? Compute each $T_{c} / M$ .

Step 3. Ratio $= 132/150 = 0.88 = 0.938$ . Why this step? $v_{e}$ drops to $93.8%$ — a $6.2%$ loss of exhaust speed despite the hotter chamber.

Verify: $150 > 132$ so ratio $< 1$ ✓, confirming the counter-intuitive result: heavier products beat the extra heat. This is why optimum $O / F$ is slightly fuel-rich — the parent's mistake box, now quantified. ✓

Ex 8 — Cell I · cross-compare with cryogenic

Worked example Same Δv, which needs less propellant?

To achieve $Δ v = 4000 m/s$ with a $500 kg$ dry stage, compare N₂O₄/MMH ( $I_{s p} = 320 s$ ) with a LOX/LH₂ stage ( $I_{s p} = 450 s$ ). How much propellant does each need?

Forecast: The higher- $I_{s p}$ cryogenic stage should need less propellant. By how much — half, a bit less?

Step 1. Invert Tsiolkovsky: $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}}$ , so $m_{0} = m_{f} e^{Δ v / (I_{s p} g_{0})}$ . Why this step? We know Δv and want mass; rearrange, using $ln$ and $e$ as inverse operations.

Step 2 (hypergolic). $v_{e} = 320 \times 9.81 = 3139.2$ ; exponent $= 4000/3139.2 = 1.2742$ ; $m_{0} = 500 e^{1.2742} = 500 \times 3.576 = 1788 kg$ . Propellant $= 1788 - 500 = 1288 kg$ . Why this step? Plug in.

Step 3 (cryogenic). $v_{e} = 450 \times 9.81 = 4414.5$ ; exponent $= 4000/4414.5 = 0.9061$ ; $m_{0} = 500 e^{0.9061} = 500 \times 2.4747 = 1237 kg$ . Propellant $= 1237 - 500 = 737 kg$ . Why this step? Same recipe, higher $v_{e}$ .

Verify: Cryogenic needs $737$ kg vs hypergolic $1288$ kg — about $43%$ less ✓, consistent with its higher $I_{s p}$ . But recall: cryogenics boil off and can't wait years — the whole reason Solid Rocket Propellants and hypergolics exist. ✓

Recall Which cell was which? (self-test)

"Doubling propellant only adds a fixed 2176 m/s" — which cell? ::: Cell B, the large-ratio $ln$ limit (Ex 2). "Burning 0 kg gives Δv = 0 because ln 1 = 0" — which cell? ::: Cell D, the degenerate/zero input (Ex 3b). "Hotter chamber but lower I_sp" — which cell and why? ::: Cell H; because $v_{e} \propto T_{c} / M$ and heavier products win (Ex 7). "40 pulses give the same Δv as one burn" — why? ::: The logs telescope; only start and end mass matter (Ex 6, cell G).

Connections

Tsiolkovsky Rocket Equation — the engine behind Ex 1, 2, 3, 6, 8.
Specific Impulse — converting $I_{s p}$ to $v_{e}$ in every example.
Reaction Control Systems (RCS) — the pulse-train problem, Ex 6.
Arrhenius Rate Law — the ignition-delay exponential, Ex 5.
Combustion Thermodynamics — the $T_{c} / M$ trade-off, Ex 7.
Cryogenic Propellants — LOX-LH2 — the cross-comparison, Ex 8.
Solid Rocket Propellants — the other storable option.