Where it comes to fuels, Specific Energy and Energy Density are two major quantities deciding "how good" given fuel is. These are either given neglecting the oxidizer mass, or as a value combined with the oxidizer, as bipropellant.

What would be the corresponding quantity measuring efficiency of oxidizers? E.g. if the rocket uses the same amount of the same fuel, what measure would tell how its energy output to mass ratio changes when switching between LOX, nitrous oxide or peroxide - each in stochiometric ratio to the fixed amount of fuel? (...or some other method of measurement that yields a similar knowledge).

Some table with example values for various oxidizers would be nice too.

  • 2
    $\begingroup$ I'm not a chemist, but maybe (1) you can only compare the energy density of fuel/oxidizer combinations, not fuels in isolation and (2) the oxidizer is implicitly assumed to be stochiometric oxygen, so it appears that fuel is being discussed rather than the combo. $\endgroup$ – Russell Borogove Jun 4 '16 at 22:30
  • $\begingroup$ @RussellBorogove: I might get some of them, but I wish I had at least keywords to look for... In particular, for amateur rocketry, I'm interested in comparing nitrous oxide and potassium nitrate; for same amount of fuel, which oxidizer will produce more delta-v? Sure I could just find mass of each needed to produce a mole of oxygen, but that doesn't take into account the bonds of oxygen with these substances - energy required to free the oxygen. Also, potassium nitrate rarely shows up in context of rocket propellants. I'd ask about that. But first I want to know what should I be asking about! $\endgroup$ – SF. Jun 5 '16 at 0:37

A good way to get a feel for the relative strengths of various oxidizers is to compare their standard reduction potentials. Fluorine is one of the most potent oxidizers known, with a reduction potential of +2.87 V. Here are some others that you might find relevant:

S$_2$O$_8$$^2$$^-$, +2.01 V
H$_2$O$_2$, +1.78 V
MnO$_4$$^2$$^-$, +1.70 V
Cl$_2$, +1.36 V
O$_2$, +1.22 V
NO$_3$$^-$, +0.96 V

Bear in mind that these potentials are really showing the oxidizer's ability to oxidize water (or protons), but their order should remain the same regardless of what molecule is being oxidized. Proper choice of an oxidizer for a rocket propellant would take into consideration both the oxidizer's strength and the mass required to combust a given amount of fuel.

| improve this answer | |
  • $\begingroup$ So, potassium nitrate at 101g/mol and liberating one oxygen atom, N2O 44g/mol, one atom; O2 32g/mol two atoms, that is 16g/mol of oxygen. That would be the mass:oxygen content data, producing the same amount of energy; still this doesn't take into account energy of the original bonds that need to be broken in order to liberate the oxygen - how would that look like? $\endgroup$ – SF. Jun 5 '16 at 10:42
  • $\begingroup$ I don't think you will be able to get actual values for the free energy released when different oxidizers react with your fuel. That would require combustion experiments in a calorimeter. In the absence of such data, about the best you can do is pick the strongest oxidizer with the lowest mass requirement. Although, it could very well be possible that a very strong oxidizer with a higher mass requirement might outperform a weaker oxidizer with a lower mass requirement due to the extra energy that is released upon combustion. $\endgroup$ – Qubit1028 Jun 5 '16 at 12:48
  • $\begingroup$ I don't have access to a lab to perform such experiments, especially a calorimeter capable of containing rocket fuel combustion reaction. I'm fairly sure these values are available somewhere but I don't even know what exactly to look for. Plus reaction energy is not an ultimate measure. There are secondary factors, e.g. physical properties or safety of the oxidizer. I'd outright disregard potassium nitrate if it wasn't easier to handle than nitrous oxide (SRB vs hybrid motor complexity). OTOH currently potassium permanganate looks far more promising in the solid oxidizers domain. $\endgroup$ – SF. Jun 5 '16 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.