10
$\begingroup$

So for instance, for the space shuttle,liquid hydrogen will react with liquid oxygen to produce water vapour. Is only water vapour ejected from the rocket? Or will the ejected mass be a mix of water and hydrogen?

If the ejected mass is a mix, is it done on purpose? Or it is not possible to eject purely water even with the right stoichiometric ratio? What is the desired mix then?

$\endgroup$
6
  • 3
    $\begingroup$ Ideally you'd only want water vapour to be ejected from the rocket, as ejecting unreacted hydrogen is wasting precious energy. $\endgroup$ Commented Jan 4, 2018 at 7:46
  • 7
    $\begingroup$ actually it's more efficient to eject hydrogen and steam due to exhaust velocity being higher $\endgroup$
    – user20636
    Commented Jan 4, 2018 at 8:15
  • $\begingroup$ It might be necessary to eject hydrogen and steam to protect the rocket engine for oxidation. Very hot metals and steam tends to react producing metal oxides and hydrogen from the steam. $\endgroup$
    – Uwe
    Commented Jan 4, 2018 at 9:14
  • $\begingroup$ Related (but not a complete answer): space.stackexchange.com/questions/22148/… $\endgroup$
    – Hobbes
    Commented Jan 4, 2018 at 10:54
  • $\begingroup$ Also related: space.stackexchange.com/questions/22122/… $\endgroup$ Commented Jan 4, 2018 at 18:06

2 Answers 2

11
$\begingroup$

In rocket engines, the objective of the combustion is to produce the maximum ISP, which is to say that the gasses exiting the nozzle have the highest possible speed.

In a typical scenario, changing the chamber pressure does not influence the ISP much. However, the exit velocity is influenced by the molecular weight of the gases, since the same exit Mach number means a higher speed for a light gas than it means for a heavy gas.

This is part of the optimization, which should ultimately lead to optimal performance parameters of the rocket, not the engine, while also taking other factors into account.

For hydrogen, stoichiometric is at a ratio of 8:1, and optimal ISP is attained at a mixture ratio as low as 3.5:1, meaning that more than half the hydrogen is ejected unburnt. This would lead to extremely large and heavy hydrogen tanks, however, so an optimal choice will be something in between.

For hydrocarbon fuels, the same is true to a lesser extent. CO2 is a quite heavy molecule, so rockets also eject a lot of CO, in order to obtain higher ISP.

$\endgroup$
5
  • $\begingroup$ I don't think that exit velocity matters; I think exit momentum of the molecules/particles matters. $\endgroup$
    – Crowley
    Commented Jan 4, 2018 at 16:44
  • 2
    $\begingroup$ The objective is not to produce the maximum specific impulse. It is to minimize the cost of getting a payload into orbit, or perhaps to maximize the payload mass. This might well mean using a suboptimal specific impulse. $\endgroup$ Commented Jan 4, 2018 at 18:15
  • $\begingroup$ I think the statement that "the objective of the combustion in a rocket engine is to expel the gasses at maximum speed" is valid as it stands, since I clarify this further down. $\endgroup$ Commented Jan 4, 2018 at 23:29
  • $\begingroup$ The objective of a rocket engine isn't always to maximise ISP. For the first stage, thrust to weight is far more important. An extreme example is ion thrusters which have massive ISP, but very little thrust. If ISP was the most important thing, then we would use ion thrusters in our first stages. $\endgroup$ Commented Jan 5, 2018 at 0:25
  • $\begingroup$ The objective of the combustion itself (not the engine) is still to accelerate the gases as much as possible. $\endgroup$ Commented Jan 5, 2018 at 2:04
5
$\begingroup$

Basically they do testing on the ground to determine the optimal mixture, and use that mixture. This website shows a few of them, of which I will post one chart here. Note that these are mass ratios.

enter image description here

So you figure out what your chamber pressure should be, and then you mix the gasses as is optimal. That will often not be a pure mixture of the two elements to do a complete burn. For LOX/ LH, the optimal point will have some unused fuels. For other fuels, the mixture may favor a different mixture as the pressure increases in general.

$\endgroup$
5
  • $\begingroup$ I remember hearing that usually a "fuel-rich" mixture is used, IIRC since the accelerated unburned fuel components slightly increase the thrust. Do you have some infos about this as well? $\endgroup$
    – DarkDust
    Commented Jan 4, 2018 at 11:32
  • $\begingroup$ Added a bit more about that kind of stuff. $\endgroup$
    – PearsonArtPhoto
    Commented Jan 4, 2018 at 11:37
  • 1
    $\begingroup$ The stoichiometric ratio is 8:1 and and a mixture ratio of the 6:1 is hydrogen rich, not oxygen rich, see. May be something is wrong here? For what fuel oxidizer combiination the diagram is valid? $\endgroup$
    – Uwe
    Commented Jan 4, 2018 at 11:50
  • $\begingroup$ You are right, sigh... That's what I get for doing rocket science first thing in the morning! $\endgroup$
    – PearsonArtPhoto
    Commented Jan 4, 2018 at 13:21
  • 2
    $\begingroup$ @DarkDust, maximum specific impulse is obtained when the exhaust has the lowest molecular weight possible. If the fuel is lighter than the oxidizer (eg. hydrogen/oxygen), you want unburned fuel. If it's the other way around (eg. kerosene/oxygen), you want unreacted oxidizer (or, in practice, partially burned fuel in the form of carbon monoxide). $\endgroup$
    – Mark
    Commented Jan 4, 2018 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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