Energy-to-weight ratio is an important parameter. We would like it to be high so we don't have to carry around much fuel.

Maybe you've heard some of the numbers for common fuels. Hydrogen is about 120 MJ/kg, diesel around 43 MJ/kg, ethanol around 28 MJ/kg, etc. However, these do not take into account the mass of the oxidizer. They only include the mass of the fuel.

That might be okay when you get free oxygen in the air, but for rocket engines, we need to carry the mass of the oxidizer with us. It sucks because the oxygen atom is fairly heavy, but we must do it nonetheless in order to make calculations based on reality. Therefore, we need to redo the math and ask, what are the energy densities of some fuels when including the oxidizer?

P.S., I wanted to ask this question even though I'm answering it myself, because I found some fairly surprising results that I think are worth sharing.


2 Answers 2


First we have to go back to the chemical equations, and this time, include the standard enthalpy of combustion.

Hydrogen: 2 H$_2$ + O$_2$ → 2 H$_2$O + 572 kJ/mol

Methane: CH$_4$ + 2 O$_2$ → CO$_2$ + 2 H$_2$O + 889 kJ/mol

Dodecane: 2 C$_{12}$H$_{26}$ + 37 O$_2$ → 24 CO$_2$ + 26 H$_2$O + 15,026 kJ/mol

Ethanol: C$_2$H$_5$OH + 3 O$_2$ → 2 CO$_2$ + 3 H$_2$O + 1371 kJ/mol

Ammonia: 4 NH$_3$ + 3 O$_2$ → 2 N$_2$ + 6 H$_2$O + 1267 kJ/mol

Carbon 1: C + O$_2$ → CO$_2$ + 394 kJ/mol

Carbon 2: 2 C + O$_2$ → 2 CO + 567 kJ/mol

And I'll just add the values for atomic weights that I'm using. These are in grams per mole. I got them off Wikipedia's Periodic Table, the big one.

H: 1.008

C: 12.011

N: 14.007

O: 15.999

From this we can calculate the energy-density by getting the mass-per-mole of one side of the equation (doesn't matter which side, since the equations are balanced), and the heat energy per mole from the end of the equation. Divide energy per mole by mass per mole and you will get the energy density.

For example, hydrogen. 4*1.008 + 2*15.999 = 36.03. 572/36.03 = 15.876 kJ/g which is equivalent to 15.876 MJ/kg.

I decided to do this for both with and without O$_2$, to see them side by side.

Fuel $\hspace{2.0cm}$ without O$_2$ $\hspace{1.7cm}$ with O$_2$

Hydrogen $\hspace{1cm}$ 141.865 MJ/kg $\hspace{1cm}$ 15.876 MJ/kg

Methane $\hspace{1.2cm}$ 55.414 MJ/kg $\hspace{1.2cm}$ 11.107 MJ/kg

Dodecane $\hspace{0.9cm}$ 44.106 MJ/kg $\hspace{1.3cm}$ 9.856 MJ/kg

Ethanol $\hspace{1.4cm}$ 29.760 MJ/kg $\hspace{1.25cm}$ 9.651 MJ/kg

Ammonia $\hspace{1.05cm}$ 21.456 MJ/kg $\hspace{1.25cm}$ 8.172 MJ/kg

Carbon 1 $\hspace{1.1cm}$ 32.803 MJ/kg $\hspace{1.25cm}$ 8.953 MJ/kg

Carbon 2 $\hspace{1.1cm}$ 23.603 MJ/kg $\hspace{1.2cm}$ 10.121 MJ/kg

(P.S., these are the higher-heating values (HHV). The lower-heating values exclude energy carried away by vaporized water. I think for rocket engines, we want HHV because that carried away water still plays a role imparting an impulse to our vehicle, via Newton's 3rd law.)

There are some interesting things about this. First thing I noticed was that Hydrogen is really not all that it's cracked up to be. When including the mass of the O2, the drop away from hydrogen is not all that significant compared to the usual MJ/kg figures we're used to. Ammonia + O2 still has over 50% as much energy as hydrogen + O2!

Second thing I noticed was ethanol vs dodecane. Dodecane is basically the highly pure kerosene used in RP-1 (the Russian version is called T-1 I think). But with oxidizer mass included, ethanol and dodecane are almost exactly equal! Maybe von Braun was not so primitive to use ethanol in his V2 after all?

Third thing, and this one really blew me away. Look at Carbon equation 2. That's the one that produces carbon monoxide instead of carbon dioxide. (Carbon monoxide is produced when there is "insufficient" oxygen.) Guess what. Burning it this way produces more energy! You just have to take into account the mass of all reactants, like rocket engineers should. This result is so surprising that I confess I'm pretty suspicious of it. I will look for a corroborating source ot make sure I got the combustion enthalpy correct. If it's true, then well if only we could get pure carbon solid fuel working in a rocket engine... It would even beat kerosene by a little bit!

I know this is not the whole story, so it won't explain everything. Volumetric density, toxicity, temperature (cryogenic!), solid residue like soot that can wreck your turbopump... I'm sure there are other factors too. But I hope you will at least agree that this sort of calculation, taking the mass of the O2 into account, absolutely had to be done for rocket engines. Rockets carry the oxidizer with them, after all. The results are interesting.

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    $\begingroup$ Is there a reaction pathway at reasonable chamber pressures and temperatures that would preferentially produce CO instead of CO2? I know of some reactions that produce CO but at considerably higher pressures. One possible (undesirable) result of trying to do this is that supplying oxygen in the stoichiometric ratio to produce CO winds up producing CO2, a little CO, and a lot of unreacted (or reacted with something else) C, maybe in the form of "soot" (Buckyballs, etc.) $\endgroup$ May 14, 2018 at 6:25
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    $\begingroup$ At high temperatures 2CO is more stable than C + CO2 because of the entropy of the extra mol of high temperature gas. $\endgroup$ May 14, 2018 at 7:17
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    $\begingroup$ Except none of this is very useful for building rockets. ISP is far more important than energy content. The products of the reaction have a huge effect on your thrust. Lighter exhaust gases will be accelerated to higher average velocities given the same temperature. Internal structure of the gases can rob is of valuable energy by giving us additional modes of excitation...etc... $\endgroup$
    – Aron
    May 14, 2018 at 10:08
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    $\begingroup$ Von Braun used some water in the ethanol to limit combustion temperature to make chamber cooling easier. Ethanol may be mixed with water, but not dodecane. Ethanol was made from potatos and did not use the limited resources of oil or coal during WWII. $\endgroup$
    – Uwe
    May 14, 2018 at 16:07
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    $\begingroup$ @Aron exhaust molecule size actually cancels out of the equation when starting with energy per unit mass and deriving max theoretical exhaust velocity. The reason molecule size matters is temperature. Smaller molecules have a higher molecular velocity at the same temperature. Since engine materials have temperature limits, lighter molecules make better propellants. $\endgroup$
    – Lex
    May 16, 2018 at 6:34

I think for rocket engines, we want HHV because that carried away water still plays a role imparting an impulse to our vehicle

It very much does play a role, but still, just taking the HHV isn't very sensible here. The HHV is specifically the energy you get directly from the reaction to gaseous products, plus the energy you can extract by condensing the water back to the liquid phase. ...Which can certainly be done in efficient heating or power applications, making the HHV a sensible metric, but it can not be done in a rocket. Again, the water vapour energy is still used for propulsion, but so is the energy in the carbon dioxide. So wouldn't you need to add also the energy that could be extracted by condensing that $\mathrm{CO_2}$ to dry ice?

No, in fact neither. As far as energy is concerned, only the LHV is relevant in a rocket. However, energy alone buys you nothing in a rocket. For an extreme example: if you tried to fuel your rocket with liquid sodium and chlorine, the reaction product would be salt. Try to expand that in a nozzle, and it will just condense to crystals. No propulsion accomplished at all.

What you really need is mechanical energy via pressure, and that you get only by using the energy to expand some gas. This basically follows the ideal gas equation $$ P\cdot V = n\cdot k\cdot T $$ The temperature $T$ is what's raised by energy from the reaction. How much it is raised depends also on the adiabatic exponent, but like the Boltzmann constant $k$ we can consider that as constant for the purpose of this discussion.

What's not constant though is the amount of substance $n$. This is just a count of molecules, so expanding many light water molecules gives you much more pressure and thus thrust than using the same energy to expand fewer $\mathrm{CO}$ or even $\mathrm{CO_2}$ of the same mass.

An alternative and arguably more to-the-point way to look at this is to consider the velocity of the gas after expansion. Light water molecules attain a higher speed with the same kinetic energy.

Thus, as Aron commented, hydrogen-rich reactions give you more benefit that your figures suggest in terms of specific impulse, which is ultimately the single most important quantity to rate how efficient a rocket is.

Your estimations are still correct in the sense that hydrogen is not many times more efficient than carbon-containing fuels. However, ever a small looking factor better $I_\mathrm{sp}$ can buy you a vastly better payload/fuel ratio, because $I_\mathrm{sp}$ enters exponentially in the rocket equation.

You could use the energy from the $\mathrm{NaCl}$ reaction to heat some other gas, such as pure hydrogen. But you'd have to carry that separately then, which is silly compared to just using it in the reaction. (If the energy comes from something else but a chemical reaction this does make sense though, notably in a nuclear thermal rocket.

The adiabatic exponent depends on how many microscopic degrees of freedom the molecules have. For a diatomic gas, these are just the rotation modes (which is a point in favour of $\mathrm{CO}$ or $\mathrm{N_2}$), for large molecules it also includes vibration modes (which makes anything with more than three atoms pretty useless).

  • $\begingroup$ There's an equation for effective exhaust velocity. A giant square root, and inside is something like TR/M * (Pa - Pe)^(k/1 - 1). Something like that. Strangely, I cannot find it now. If you know it, please post it. I want to calc theoretical max Ve and add it to my chart, to see a comparison. One more thing. Never understood how you can go from chemical energy to Ve alone (the Tsiolkovsky equation uses Ve in m/s instead of ISP in seconds)...cuz kinetic energy has v squared. Maybe you can add something about that, since for some reason the Tsiolkovsky equation doesn't have squares or sqrt. $\endgroup$
    – DrZ214
    May 14, 2018 at 16:09
  • $\begingroup$ Effective exhaust velocity is just (Thrust * g / weight flow rate). (Sutton equation 2-22, 4th edition) $\endgroup$ May 14, 2018 at 17:00
  • $\begingroup$ @OrganicMarble Yes but then how do you calculate thrust based on temperature, pressure, molecular species, etc? I found it. braeunig.us/space/index.htm Equation 1.22. Stupidly, it is far, far below equation 1.2 because that website pretends that fractions are integers. 1, 2, 3 ... 21, 22. Anyway, there are many ways to calculate something. I mentioned Ve, you mentioned thrust. I go with Ve because that's what's used in the Tsiolkovsky equation. I will use it and add to my chart soon. $\endgroup$
    – DrZ214
    May 15, 2018 at 4:47

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