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Hydrogen combustion net momentum

  1. May 21, 2007 #1
    Hey what if you had a certain volume of hydrogen gas in two different scenarios. They are of equal amounts. In one scenario, the hydrogen is ignited so that it explodes instantaneously. In the other scenario, the (equal amount of) hydrogen is used in a thrust engine for an extended period of time.
    Is the net momentum (F x dt) the same in both situations?
     
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  3. May 21, 2007 #2
    I don't thinks so.
    When hydrogen explores immediatelly, its net momentum is zero due to its symetry. And if it explores in a thrust engine for extended period of time, the net momentum is not zero because it is asymetric.
     
  4. May 21, 2007 #3

    russ_watters

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    I suppose the OP is thinking of an explosion pushing on a pusher-plate. Such a device is a very inefficient way of capturing and directing the energy of combustion.
     
  5. May 22, 2007 #4
    pusher-plate? do you mean like in a piston?
     
  6. May 22, 2007 #5

    russ_watters

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    The only means of using explosions as propulsion that I've ever heard of was with the Project Orion nuclear-bomb powered rocket. The bottom of the rocket was a large shallow-curved plate and the bomb exploded right below it, pushing the rocket up.

    If that isn't what you were getting at.....what were you getting at?
     
  7. May 22, 2007 #6
    I dont understand what you said. I think that we should concentrate on the main problem (the net momentum), not its efficiency. ok?
     
  8. May 22, 2007 #7

    russ_watters

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    What I mean is efficiency of utilizing the energy - a pusher-plate is essentially just a very bad nozzle. The OP wasn't very specific about what this is all about, though, so it is tough to know the real intent of the question. You were right that an explosion in the open has zero net momentum, so the OP question wouldn't make a whole lot of sense to be asking about that.
     
  9. May 23, 2007 #8
    Hey russ_waters, sorry for being thick-headed, but could you give me an example of a pusher-plate?

    Well what I was getting at, or what I am trying to do, is to use eddy currents (magnet and copper pipe) to "transform", if you will, the impulse of an explosion. That is why I was wondering if an explosion offers the same change in momentum in one direction as plain old thrust (for the same given amount of time that the explosion was active of course).

    Just to clarify, what I say "transform" in the electrical transformer sense. Change in momentum is F x dt. Call Voltage F, and Amperes dt. Suppose explosion has high V but low A. Eddy currents will lower the V (instantaneous force) but extend A (dt).

    Just another thought that I really have no confidence in, but where is the momentum advantage of a (heat-wasting) hydrogen car over plain old thrust? Or is that two totally different ball fields?
     
  10. May 23, 2007 #9

    russ_watters

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    The top-right picture is the project Orion spacecraft sketch: http://www.daviddarling.info/encyclopedia/O/OrionProj.html

    I'm still not clear about what you are trying to do - when you talk about cars, are you now talking about burning hydrogen in a regular internal combustion engine?

    Chemically, there is only one possible amount of energy released in a chemical reaction, but the nitty-gritty of how it is harnessed is what determines how much useful energy/change in momentum you get from it.

    Also, momentum ("net" or otherwise) and change in momentum aren't the same thing. That's part of what didn't make sense in your first post.
     
  11. May 23, 2007 #10
    The question is not clear, howver, if your mean total change of momentum due these two different processes, where total change of momentum is defined as [tex]\Delta P=P_f-P_i[/tex], then in an ideal case they are the same due to "there is only one possible energy release"(russ_watters). But "nitty-gritty of how it is harnessed is what determines how much useful energy/change in momentum you get from it" (russ_watters).
     
  12. May 24, 2007 #11
    Oh sorry about the use of "net" early in the post. I meant change. I ll change it.

    Yes I mean h2 car.
     
  13. May 24, 2007 #12

    russ_watters

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    Ok, that makes the question a lot simpler. An internal combustion car engine runs at around 35% thermodynamic efficiency. Nozzles can be upwards of 90%.
     
  14. May 24, 2007 #13
    wow that's pretty high, i suppose the reason they dont do that in cars is because of the car behind!

    you've got a point there, this wouldnt be too hard to put inside a car as an engine would it?
     
  15. May 24, 2007 #14
    Hey russ_waters, I knew the heat efficiency of nozzles were much higher than the combustion engine cars, but can you compare them in terms of change in momentum per a given amount of hydrogen?
     
  16. May 25, 2007 #15
    If I have one device that uses the eddy current impulse "transformer" stated earlier (w/ explosions of hydrogen), and compare it to a rocket engine with thrust to equal the amount of hydrogen used by the explosion during any given amount of time, the force and change in momentum provided by the two will be the same, correct?
     
  17. May 26, 2007 #16
    only if they have the same efficiency...

    in a perfect world, yes.

    if efficiency is lower, more of the energy from the hydrogen combusting is converted into useless byproducts (eg heat). I dont understand the eddy currents thing, but as that and the rocket seem to be different devices, they will probably have different efficiencies.

    is this eddy impulse thing what is supposedly being used on the aurora?

    also friction is important... i think the air resistance that both encounter would overall be the same. the slower burning rocket would have a steadily increasing air resistance, whereas your eddy current impulse rocket thing would experience brief accelerations to a higher speed, therefore it would not experience the air resistance between the two speeds as much as the slow burner... but it would then be exposed to a larger air resistance for a time until the next explosion... therefore, i think the same speed is achieved in both... so i think air resistance can be ignored as with identical efficiencies i think the same speed would be achieved.

    so yh i think you are right but only aslong as efficiencies are the same...
     
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