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Rocket Propellants Problem

  1. Oct 26, 2013 #1
    So, as I pet project I'm designing things, and I've come across an issue:
    I'm trying to design propellant tanks and such for a vehicle, so I need to find the mix ratio of the propellants. In this case, it's Liquid Oxygen and Methane.

    CH4 + 2O2 → CO2 + 2H2O

    Using atomic masses and such, it would seem the best mix ratio would be 2:1, by mass, oxidizer to fuel.

    However, on Encyclopedia Astronautica, consistently the engines with the highest efficiency (specific impulse) have an Oxidizer to Fuel ratio of 3.4
    (big list of engines using these propellants)

    So, I decided, maybe it's by volume.

    So, by volume, just plugging in the densities, where LOX has 1.14 g/cm3, and CH4 has 0.424 g/cm3,

    1 / 1.14 : 1 / 0.424
    = 1 : 2.688
    Which still doesn't line up with O:F = 3.4

    Any help appreciated, thanks.
  2. jcsd
  3. Oct 26, 2013 #2


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    You need to have enough oxidizer for the fuel; this usually means that you need a surplus of the oxidizer - you certainly don't want to eject unused fuel!

    Now the most efficient might mean effective propulsion, which in turn depends upon the pressure, temperature, and chemical ratios, as well as the geometry of the rocket.

    I would expect that the ratio 3.4 is empirical: you have to run tests for your design.

    Also I suspect it is the ratio of volumes at the working temperatures (cryogenic), which then relates directly to how much you have to pump.
  4. Oct 26, 2013 #3
    The given densities are at the cryogenic temperature ranges they're used at. I can also see how it'd be off a little - and I believe they try to run them slightly fuel rich, since unused oxidizer is just as bad as unused fuel, except extra oxidizer will corrode parts more than unused fuel, but going from 1 : 3 to 3.4 seems right out of the ballpark.

    Also, the Space Shuttle ET carried Oxidizer to Fuel at about a 6:1 ratio, [source]

    2H2 + O2 → 2H2O
    Which is 8:1 by mass.

    Encyclopedia Astronautica reports it as 6.
    Okay, I think my big issue is actually a misunderstanding in reading it. 3.4 denotes 3.4:1, not 3:4. Well that's a derp, haha. derpyhoovesliarplz.png

    And it looks like it is by mass, since with the SSME's it lines up with what wiki says on the ET. 3.4:1 is quite a bit off but not too far off from 2:1.

    Still seems a bit astounding that that much efficiency is wasted, though. That's a lot of unburnt fuel! Though perhaps there's secondary reactions in the combustion chamber and that CO2 ends up getting burnt apart, as well; I have heard of CO2 being used as a propellant. Not sure what that chemical reaction would look like, though, I do have a bit of a weak spot in chemistry.
    Last edited: Oct 26, 2013
  5. Oct 26, 2013 #4

    D H

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    In many cases, that is *exactly* what you want do. Consider an extreme case, LOX/LH2. The stoichiometric ratio is 8:1 oxygen:hydrogen by mass. Hydrogen is so light that it makes more sense to run extremely rich. Ignoring that more fuel means a bigger fuel tank, somewhere between 4:1 and 4.5:1 ratio would be ideal. Taking that bigger fuel tank into account means running not quite as rich as that, but still rather rich. The Shuttle, for example, used a 6:1 mixture ratio.

    As a general rule, you should run rich if the fuel molecules are lighter than the oxidizer molecules, lean if it's the other way around. A stoichiometric ratio will optimize energy output per unit mass of fuel+oxidizer mix. That's not the goal with rockets. The goal is to optimize thrust, or change in momentum, per unit mass of fuel+oxidizer mix.

    The reaction for LOX/methane is 2O2+CH4→CO2+2H2O. As a molecule of O2 has an atomic mass of about 32, and for methane it's about 16, that calls for a stoichiometric ratio of 4:1 oxygen:methane. However, since methane is lighter than oxygen, you're better off running slightly rich, somewhere between 3:1 and 3.5:1.
  6. Oct 26, 2013 #5
    Well put, but that does leave one question. Why isn't optimizing energy/mass the same as optimizing thrust/mass, since that thrust comes from kinetic energy from the combustion?
  7. Oct 26, 2013 #6
    This is not about side reactions. Quite simply, fuel-rich mixtures are used to maximize the exhaust velocity.

    You may have the misconception that if you release the maximum amount of energy in combustion, then you will get the maximum exhaust velocity. That is not the case, because some of that energy will be released in the form of heat, which is useless for propelling the rocket, and the amount of wasted energy depends, among other things, on the composition of the exhaust.

    The thermodynamics is complex, and I'm pretty sure the rocket scientists can do it better than I can, so if they say a 6:1 oxygen-hydrogen ratio is optimal for the SSME, then it probably is.

    EDIT: Your most recent post came while I was writing this. As I said above, the energy that you get out of the fuel is divided into two components -- heat, which does not help to propel the rocket, and kinetic energy, which does. It's quite possible to make a change to the mixture ratio that reduces the total amount of energy, but increases the kinetic energy.
  8. Oct 27, 2013 #7

    D H

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    Short answer:
    It's because of thermodynamics, the kinetic theory of gases, and chemistry.

    Not-so-short answer:
    The kinetic theory part is perhaps the easiest to understand. Kinetic theory says that the square of the mean velocity of a gas molecule is proportional to temperature, inversely proportional to the number of degrees of freedom and inversely proportional to the molecular mass. Consider LOX/LH2. Oxygen and hydrogen burn to form water, a polyatomic molecule. Burning rich reduces the degrees of freedom and reduces the mean molecular mass, but at the cost of a decrease in flame temperature. Obviously there's a trade-off here. Burn way too rich and the losses due to the drop in flame temperature overwhelm the gains due to the reduction in degrees of freedom and the reduction in molecular mass. With LOX/LH2, the optimal point in terms of thrust is at about half of the stoichiometric ratio.

    Now consider your fuel, LOX/methane. Ideally, LOX and methane at a stoichiometric ratio burn to form carbon dioxide and water. Not so ideally, CO2 at high temperatures has a nasty propensity to dissociate to carbon monoxide and oxygen, which is an endothermic reaction. This dissociation quenches the flame temperature a bit. You'll never achieve that stoichiometric ideal even if the ratio is stoichiometric. Burning a bit rich will naturally yield carbon monoxide in favor of carbon dioxide. Burning at a 3:1 ratio would (ideally) be equivalent to 3O2+2CH4→2CO+4H2O: No CO2! That's the ideal; reality is never ideal. Instead you'll get some unburnt methane in the exhaust at this ratio, which is not good at all. In practice, the best ratio is somewhere between 3:1 and 3.5:1, depending on how well your thruster burns things and on the ambient pressure.

    The above over-simplified things. The kinetic theory of gases is just a model. There is no such thing as an ideal gas, and at these temperature, gases are not ideal. The real answer is found empirically.
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