Several questions and answers here speak of stochiometric ratio. I do understand it is convinient for efficiency (greater ISP). I also read somewhere (I am unable ro find again the reference, it was not that one) that it is not uncommon to run a fuel-rich combustion in the combustion chamber (i.e. less oxydizer than in stochiometric ratio), mainly for cooling reasons.

My question is mainly asked because I wonder why stochiometric ratio seems important whereas fuel-rich is commonly implemented. Note that I have no idea of how rich is fuel-rich.

  • $\begingroup$ The stochiometric ratio doesn't necessarily give the best efficiency. None of the three questions/answers found by the search seem to assign any particular importance to it. $\endgroup$
    – user20636
    Apr 4, 2018 at 11:43
  • $\begingroup$ This answer seems pretty much to address the issues -- space.stackexchange.com/questions/22122/… $\endgroup$ Apr 4, 2018 at 13:45

2 Answers 2


There are several factors as to why an engine would not run the “ideal” stoichiometric ratio of fuels. Here are the two main reasons why rocket engines wouldn’t run the “ideal” ratio.

Higher exhaust velocity

An equation defining exhaust velocity in Richard Nakkas website shows that molecular mass of the exhaust products ends up on the denominator of the equation. This means that the lighter the exhaust product is, the faster the exhaust velocity. So it may be beneficial to run the engine not stoichiometrically ideal in order to keep the molecular mass the least. This is why most hydrolox engines run fuel rich as hydrogen is much lighter than oxygen, making for a more efficient engine as combustion products contain much more hydrogen.

Lower combustion temperature

While that same equation defined in the above paragraph shows that a higher temperature improves the efficiency, too high of a temperature can lead to the engine becoming damaged and the process of rapid disassembly will occur. This is why some engines are run rich in either fuel of oxidizer to ensure that the combustion temperature is certainly something that the chamber and nozzle can handle.

  • 2
    $\begingroup$ Ideal in quotations as ideal stoichiometrically differs from ideal efficiency wise and ideal reliability wise $\endgroup$ Apr 4, 2018 at 15:05
  • $\begingroup$ "This is why some engines are run rich in either fuel of oxidizer" - usually fuel-rich, especially with hydrogen which having very light particles improves specific impulse (same energy accelerates particle to higher speed). Engines with excess oxidizer have a nasty tendency of running engine-rich mix. $\endgroup$
    – SF.
    Jun 19, 2018 at 21:05
  • $\begingroup$ @SF. Usually yes, but I do believe some Russian engines run oxidizer rich (kerolox engines not hydrolox) such as the Nk-33 and rd-180 $\endgroup$ Jun 19, 2018 at 21:08

As I understand it, stochiometric combustion releases the most thermal energy into the exhaust, but there are several reasons that might not be the optimal performance point.

  • Some of that thermal energy goes into molecular vibration energy rather than directed exhaust energy; you want simpler molecules with fewer vibrational modes to minimize that factor -- H2 instead of H2O, for example, in a hydrogen-oxygen rocket.

  • Running off-stochiometric means burning cooler, so you can have a simpler, lighter, or cheaper thermal system in the chamber and nozzle. (The earliest large rockets like V-2 and Redstone actually used 75% alcohol with 25% water as fuel, simply in order to run cool.)

  • By burning fuel-rich, you can ensure that a minimal amount of hot, unburned oxidizer is attacking the chamber and nozzle while the engine is running; at stochiometric ratio, if there was incomplete mixing of propellant this could cause a lot of damage.

  • If one of your propellants is significantly denser than the other, your tankage will be more compact the further you shift the mixture ratio toward the denser component. Unfortunately, in most cases the oxidizer is denser than the fuel, so this leads to an oxidizer-rich ratio, which is incompatible with the previous point.


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