Green propellants — LMP-103S, AF-M315E (vs hydrazine): Exercises

Formula The toolkit you will reuse — with every symbol spelled out

Thrust: $F = \overset{m}{˙} v_{e}$ . Here $\overset{m}{˙}$ ("m-dot", kg/s) is how many kilograms of exhaust leave per second, and $v_{e}$ (m/s) is how fast that exhaust flies out the back. Thrust is just those two multiplied — momentum thrown backward per second.
Specific impulse: $I_{s p} = \frac{v _{e}}{g _{0}}$ , where $g_{0} = 9.81 m/s^{2}$ is standard gravity used only as a unit-fixing constant. $I_{s p}$ (seconds) is a fuel-efficiency score: bigger = the chemistry throws exhaust faster.
Performance scaling: $I_{s p} \propto \frac{T _{c}}{M}$ . $T_{c}$ (K) is the chamber/flame temperature — how hot the decomposition gas gets; $M$ (g/mol) is the mean molar mass of the exhaust — how heavy the average gas molecule is. Hot helps, heavy hurts. See Thermochemistry & enthalpy of decomposition.
Density-impulse (tank metric): $ρ I_{s p}$ , where $ρ$ (kg/L) is the propellant density — how many kilograms fit in one litre. This is the number that matters when the tank volume is fixed.
Mission capability: $Δ v = v_{e} ln \frac{m _{0}}{m _{f}}$ , the Rocket equation (Tsiolkovsky). $m_{0}$ = wet mass (spacecraft with fuel loaded), $m_{f}$ = dry mass (spacecraft after fuel is spent), and $ln$ is the natural logarithm — the slow-growing function that turns a fuel ratio into a speed change.

Level 1 — Recognition

Recall Solution

LMP-103S → ammonium dinitramide (ADN), $N H_{4} N (N O_{2})_{2}$ .
AF-M315E / ASCENT → hydroxylammonium nitrate (HAN), $N H_{3} O H^{+} N O_{3}^{-}$ . Both are nitrogen-rich nitrate/dinitramide oxidiser salts: they carry their own oxygen in the $N O_{2} / N O_{3}$ groups. Mnemonic: "AF has HAN, LMP has ADN."

Recall Solution

True — with a precision caveat. LMP-103S and AF-M315E are aqueous solutions of energetic salts, not pure ionic liquids in the strict sense. What matters for safety is that the energetic species are ions dissolved in water, and ions are non-volatile — they do not evaporate into a vapour cloud the way molecular Hydrazine does. So the propellant barely puts toxic vapour into the air, which is the core safety win. The water carrier also lowers volatility. (Contrast: hydrazine's danger is largely its high vapour pressure as a volatile molecule.)

Level 2 — Application

Recall Solution

Recall $I_{s p}$ (seconds) is the efficiency score and $v_{e}$ (m/s) is the exhaust speed, linked by $I_{s p} = v_{e} / g_{0} \Rightarrow v_{e} = I_{s p} g_{0}$ . $v_{e} = 252 \times 9.81 = 2472 m/s .$ Why multiply by $g_{0}$ ? $I_{s p}$ is stored in seconds; multiplying by the constant $g_{0} = 9.81 m/s^{2}$ converts back to a true speed in m/s.

Recall Solution

Recall $\overset{m}{˙}$ is the mass leaving per second (kg/s) and $v_{e}$ is how fast it leaves (m/s). $F = \overset{m}{˙} v_{e} = 0.25 \times 2472 = 618 N .$ Why this formula? Thrust is momentum thrown backward per second (Newton's 3rd law): mass rate times how fast you throw it — exactly the left panel of the map figure above.

Recall Solution

The ratio $3 : 4 : 1$ means per $3$ mol $N_{2} H_{4}$ you get $4$ mol $N H_{3}$ and $1$ mol $N_{2}$ . Scale by $2$ (since $6/3 = 2$ ): $N H_{3} : 4 \times 2 = 8 mol, N_{2} : 1 \times 2 = 2 mol .$ See Catalysis: this happens over the Ir/Al₂O₃ (Shell 405) catalyst.

Level 3 — Analysis

Recall Solution

Recall $ρ$ is density (kg fitting in one litre) and $I_{s p}$ is the per-kg efficiency; their product is the density-impulse, the score for a fixed-volume tank. $ρ I_{s p} (LMP) = 1.24 \times 252 = 312.5, ρ I_{s p} (hydrazine) = 1.01 \times 230 = 232.3.$ $Ratio = \frac{312.5}{232.3} = 1.35 .$ Why this metric? A fixed-volume tank holds $ρ V$ kg of fuel; each kg delivers $I_{s p} g_{0}$ of impulse, so total impulse $\propto ρ I_{s p}$ . LMP-103S gives ~35% more impulse from the same tank — decisive for volume-limited small satellites.

The bar chart below makes this concrete: read the height of each bar as "impulse squeezed out of one litre of tank." Notice AF-M315E's bar towers over hydrazine's — that gap is the whole reason small satellites switched.

Figure — Green propellants — LMP-103S, AF-M315E (vs hydrazine)

Recall Solution

$\frac{T _{c}}{M} (green) = \frac{1900}{22} = 86.4 = 9.30,$ $\frac{T _{c}}{M} (hydrazine) = \frac{900}{11} = 81.8 = 9.05 .$ Green wins ( $9.30 > 9.05$ ). Why? Green's exhaust is exactly twice as heavy ( $22$ vs $11$ ), which halves the fraction — but its flame is more than twice as hot ( $1900$ vs $900$ ). Since $T_{c}$ climbed slightly faster than $M$ , the ratio $T_{c} / M$ still rises and the square root with it. Hot beats Heavy. See Thermochemistry & enthalpy of decomposition.

The quadrant map below plots each propellant with "heavier exhaust →" on the x-axis and "hotter flame ↑" on the y-axis. The diagonal contours are lines of equal $T_{c} / M$ ; green sits on a higher contour even though it is further right (heavier), because it is much further up (hotter).

Recall Solution

Catalyst light-off temperature (cold-start). Hydrazine decomposes on the catalyst at low temperature, so it starts almost instantly with little preheat. Green propellants need a much hotter catalyst preheat (high light-off T) → more electrical power and slower cold starts. For missions needing rapid, low-power cold starts, hydrazine stays relevant.

Level 4 — Synthesis

Recall Solution

Recall $m_{0}$ = wet mass (with fuel), $m_{f}$ = dry mass (fuel spent), $ln$ = natural log. (a) $v_{e} = I_{s p} g_{0} = 266 \times 9.81 = 2609 m/s$ . (b) Wet mass $m_{0} = 120$ kg; dry mass $m_{f} = 120 - 15 = 105$ kg. $Δ v = v_{e} ln \frac{m _{0}}{m _{f}} = 2609 ln \frac{120}{105} = 2609 \times 0.13353 = 348 m/s .$ Why chain these? $I_{s p}$ gives efficiency, $v_{e}$ turns it into a speed, and Tsiolkovsky turns "how much fuel + how efficient" into actual mission capability ( $Δ v$ ). The logarithm curve in the map figure (right panel) shows why the first kilograms of fuel buy the most $Δ v$ .

Recall Solution

$v_{e} = 230 \times 9.81 = 2256 m/s$ ; same mass ratio $120/105$ : $Δ v = 2256 \times ln \frac{120}{105} = 2256 \times 0.13353 = 301 m/s .$ AF-M315E gives $348$ vs $301$ m/s — about 16% more $Δ v$ from identical fuel mass, purely because its higher $I_{s p}$ raises $v_{e}$ (the $ln$ term is identical since the mass ratio is the same).

Recall Solution

Fuel mass = $ρ \times V$ (density times litres). $g_{0} = 9.81$ . (a) AF-M315E: fuel $= 1.47 \times 12 = 17.64$ kg; $m_{0} = 80 + 17.64 = 97.64$ kg; $v_{e} = 266 \times 9.81 = 2609$ m/s. $Δ v = 2609 ln \frac{97.64}{80} = 2609 \times 0.19918 = 520 m/s .$ (b) Hydrazine: fuel $= 1.01 \times 12 = 12.12$ kg; $m_{0} = 80 + 12.12 = 92.12$ kg; $v_{e} = 230 \times 9.81 = 2256$ m/s. $Δ v = 2256 ln \frac{92.12}{80} = 2256 \times 0.14113 = 318 m/s .$ AF-M315E wins massively ( $520$ vs $318$ m/s). Why the gap is bigger than L4.2? Here density and impulse both help: the denser green fuel packs 46% more mass into the same 12 L, and that extra mass makes the $ln (m_{0} / m_{f})$ term much larger too. This is the $ρ I_{s p}$ advantage (the bar chart in L3.1) made real.

Level 5 — Mastery

Recall Solution

$v_{e} = 266 \times 9.81 = 2609$ m/s. Invert Tsiolkovsky for the mass ratio (exponentiate to undo the $ln$ ): $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{200/2609} = e^{0.076656} = 1.07967.$ With $m_{f} = 90$ kg (dry mass): $m_{0} = 90 \times 1.07967 = 97.17$ kg. (a) Propellant mass $= m_{0} - m_{f} = 97.17 - 90 = 7.17 kg$ . (b) Tank volume $= \frac{mass}{ρ} = \frac{7.17}{1.47} = 4.88 L$ . Why invert? Design runs backwards: the mission fixes $Δ v$ , and you solve for the fuel needed. The exponential is the Tsiolkovsky equation rearranged.

Recall Solution

$v_{e} = 230 \times 9.81 = 2256$ m/s. $\frac{m _{0}}{m _{f}} = e^{200/2256} = e^{0.088652} = 1.09270.$ $m_{0} = 90 \times 1.09270 = 98.34$ kg → propellant $= 8.34 kg$ . Tank volume $= 8.34/1.01 = 8.26 L$ .

Verdict — compare with L5.1 (green: 7.17 kg, 4.88 L):

Green needs less fuel mass (7.17 vs 8.34 kg) and far less tank volume (4.88 vs 8.26 L — barely over half).
But the honest engineer adds: green demands a much hotter catalyst preheat → extra electrical power budget and slower cold-start (L3.3). And greens are energetic oxidisers, not inert (L1 trap) — still needing careful, though far cheaper, handling.
Overall: for a volume-limited small satellite, AF-M315E is the clear choice; for a mission needing instant low-power cold starts, hydrazine may still win. "Same job, much safer, slightly thirstier on warm-up power."

Recall One-line self-test

Why does higher density help even though it makes the loaded rocket heavier? ::: For a fixed-volume tank, higher $ρ$ means more propellant mass ( $ρ V$ ), which both adds fuel and raises the $ln (m_{0} / m_{f})$ term — more total impulse and more $Δ v$ from the same litres.