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Question about jet propulsion.

  1. May 18, 2006 #1
    In the general thrust equation in order for a jet engine, either turbojet, ramjet, or scramjet, to produce net thrust the speed of the exiting exhaust has to be greater than the incoming speed of the air flow. Then the net thrust is the mass exhaust rate times exhaust speed minus the air mass rate flow in times the air flow speed.
    But it is only the oxygen and that gets combusted with the fuel. So you would think only the oxygen portion of the air is getting accelerated. But there is 4 times as much nitrogen as oxygen.
    Is the exiting nitrogen still going to flow at the previous incoming speed thus contributing nothing to the thrust?

    Bob Clark
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  3. May 18, 2006 #2


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  4. May 18, 2006 #3


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    To continue what Astro mentioned, it is also why you see oxides of notrogen as a primary exhaust emission product.
  5. May 19, 2006 #4
    The pages you cited don't specifically say what happens to the nitrogen.
    I'm mostly interested in the ramjet and scramjet cases. For the ramjet, the supersonic air is slowed down to subsonic speeds for combustion, and for the scramjet the hypersonic air is slowed to low supersonic speeds for combustion.
    But if the larger mass of nitrogen is also thus greatly slowed down, how is the combustion of the oxygen alone going to also get the exiting nitrogen back up to above the entering speed?
    Said another way, the "ram drag" term includes the much greater mass of nitrogen. How is the oxygen alone being accelerated going to overcome the force in the opposite direction from this larger mass of nitrogen impinging on the aircraft?

    Bob Clark
  6. May 19, 2006 #5
    So you're suggesting the nitrogen is also going to be accelerated to comparable speeds to the combusted oxygen+fuel? It seems to me this can't be very efficient since there is much greater mass of nitrogen that has to be heated from conduction, convection a rather slow process.
    The degree of this mixing and heating would seem to be an important part of the thrust capability of the engine and its efficiency would depend greatly on how well this is accomplished. Curious you never see it stated specifically what this heating and velocity of the nitrogen is.
    Note this problem with a rocket doesn't arise since it is only the oxygen being combusted with the fuel producing the exhaust in that case.

    - Bob
  7. May 19, 2006 #6


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    But the combustion is only a way of generating heat, right? I mean, once the temp is raised, all the gasses expand.
  8. May 19, 2006 #7


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    Combustion is 'the' source of energy/heat in a jet engine, or any other combustion engine.

    Even with nitrogen absorbing heat and forming NOx, there is still sufficient energy and momentum for propulsion - which comes from the mass flow rate and fluid velocity.

    It is true that a rocket (using LOX and LH, or kerosene) is more efficient in the sense that it avoids the heating of nitrogen, BUT those rocket motors operate for minutes, not hours. The space shuttle main engines fire for about 8.5 minutes. The combustion chamber is running at about 6000 °F (3316°C), and the working fluid is hydrogen rich (i.e. excess H2) for greater specific impulse. The disadvantage is that one must 'carry' both fuel and oxidizer - which are rapidly depeleted and represent a fair amount of mass. The solid rocket boosters attached to the main tank are simply there to lauch the tank with the shuttle.


    Jet engines operate at much lower temperatures, so the nitrogen is not a problem with respect to propulsion, although N is problem with respect to NOx emissions.
  9. May 19, 2006 #8
    It's puzzling because whenever I see the thrust produced given for jet engines, they only give the thrust for the oxygen+fuel reaction products! For instance when they give the ISP for such engines.
    Another possibility occurs to me. When you slow down the incoming air, the nitrogen and oxygen are greatly heated and compressed. Then though they are greatly slowed from their entering speeds initially, the heating and compression allows them to exit at nearly the same speed at which they entered. That is, even if you combusted no fuel, the exiting air would still move at close to the entering speed.

    - Bob
    Last edited: May 19, 2006
  10. May 19, 2006 #9
    True but the much greater mass of the nitrogen has to be considered. This will serve to lower the overall temperature. When you light a fireplace, the entire room is not heated to the temperature of the flame.
    The lower temperature will serve to lower the exhaust speed.
    However, an interesting question occurs to me. Will the total thrust from the greater mass of the nitrogen+combustion products but at lowered temperature be the same as the thrust from the combustion products alone but at higher temperature?
    It's not obvious to me that it would. In any case if this is what happens this is something that should be mentioned in discussing the mechanism of ramjet and scramjet propulsion.

    - Bob Clark
  11. May 19, 2006 #10


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    Are you going on the assumption that the air:fuel ratio is close to stoich? The relative ratio is actually on the order of 100:1. In a typical installation, the combustion products are mixed in the dillution zone of the burner prior to entering the HP turbine to control inlet temperatures. So you have a small amount of nitrogen from the combustion process that is still being mixed with unburnt air.
  12. May 23, 2006 #11
    It seems to me the air:fuel ratio being so high would make there be even more nitrogen that is combined with the combustion products.
    I did a calculation for how much the nitrogen would degrade the performance copied below:

    Firstly, the oxygen is thorougly mixed with the nitrogen already so it
    shouldn't be a problem of applying the heat of combustion to the
    nitrogen as well.
    I also looked at a textbook that gave a calculation for a turbine jet
    engine and it calculated the thrust produced by simply applying the
    heat produced to the total air.
    However, we can estimate the loss of efficiency by the fact that the
    nitrogen is not contributing to the heat produced, i.e., it is in
    effect just being carried along.
    Let's say the temperature *increase* of the oxygen+fuel mix would be
    3000K above the temperature of the incoming oxygen if it were just pure
    oxygen being mixed with the fuel. The space shuttle main engines for
    example produce temperatures in this range. However, the space shuttle
    combustion chamber operates at 200 atm's, much higher than that of
    ram/scram -jets so our temperature increase probably won't be this
    The oxygen coming in is quite hot for (sc)ramjets because it has to be
    severely slowed down from supersonic or hypersonic speeds. Let's call
    it 1000K. Then the total temperature would be 4000K. I'll assume the
    fuel is hydrogen for simplicity.
    Now we imagine this heat is being distributed to the nitrogen as well
    as the exhaust which is now water vapor. Now the heat given to a gas is
    proportional to the temperature increase, the specific heat of that
    gas, and the mass. The specific heat of water (vapor) is rather high,
    higher than for nitrogen gas. This means it is actually easier to raise
    the temperature of nitrogen for a given heat input than water.
    However, taking into account water's higher molecular weight over
    oxygen, the mass ratio of the nitrogen to the water is about 3 to 1. So
    the total mass is about 4 times that of the water. So I'll estimate the
    temperature change for the mixture as only half. (The specific heat for
    the mixture can be calculated from those of the nitrogen and of the
    water and therefore the exact temperature increase for the mixture can
    be calculated knowing the mass is 4 times as much, but for simplicity
    I'll estimate it as about half of the pure water vapor case.) Then the
    temperature for the nitrogen+water vapor mix is 1000K + 1500K = 2500K.
    This page gives the formula for the exhaust velocity dependent on

    Combustion & Exhaust Velocity

    The formula is often written in approximate form as Ve = sqrt(2RT/M),
    where R is the universal gas constant, T is the temperature in kelvin
    and M is the molecular weight of the exhaust. So if the temperature is
    smaller by a factor of 2500/4000 and the molecular weight is larger by
    a factor of 30/18, the exhaust velocity will be smaller by a factor of:
    sqrt[(2500/4000)*(18*30) ] = sqrt(.375) = .6124

    This may seem acceptable when the mass of the exhaust is 4 times as
    high, making the exhaust thrust 4*.6124 = 2.45 times as high as the
    oxygen+fuel only case. But the drag due to the larger incoming mass is
    also 4 times as high. The thrust then is not increased by as large
    amount as is the drag.

    Bob Clark
  13. Jun 5, 2006 #12
    Hi there:

    In order to help you with the thrust numerical calculations, check out the Engineering Software free web site at: http://members.aol.com/engware/free [Broken]

    A zipped down MS Excel file is attached.



    Attached Files:

    Last edited by a moderator: May 2, 2017
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