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Maximum thrust - theory of rocket propulsion

  1. Oct 24, 2011 #1
    Hello all,

    Just wondering if anyone can give me some guidance on this: the problem is an apparent contradiction I've stumbled upon. Some sources (e.g. www.braeunig.us) suggest that maximum thrust is achieved when Pe=Pamb (i.e. exhaust pressure = ambient pressure)l whereas others say that thrust will increase as Pamb descreases (as F = [itex]\dot{m}[/itex]Ve - (Pe-Pamb)Ae). Indeed, as far as I can see, Sutton even says both! Could someone, please, help me out?!

    Many thanks,
  2. jcsd
  3. Oct 24, 2011 #2


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    Interesting cross-posting here. As I said over on the rocketry forum, if you fix the nozzle expansion ratio and vary ambient pressure, thrust is maximized for Pa = 0. However, if you fix the ambient pressure and vary the expansion ratio, thrust is maximized for Pe = Pa, and if you vary both, the overall maximum is Pe = Pa = 0. Obviously, this is unattainable in reality (as it would require an infinitely expanded nozzle), but this is why rocket engines designed to work in space have a very high expansion ratio (and a correspondingly low Pe).
  4. Oct 24, 2011 #3

    D H

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    Sutton is correct.

    A rocket converts potential energy (typically chemical) to other forms of energy. The best that can possibly be done with this potential energy is to convert every last erg to useful kinetic energy. Achieving this ideal is impossible; the second law of thermodynamics gets in the way. There are many ways in which the useful kinetic energy is less than the potential:
    • Heat in the exhaust stream will reduce the amount of energy that can be converted to kinetic energy.
    • A non-collimated exhaust stream reduces the amount of kinetic energy that is useful. (The component of velocity normal to the centerline will average out to zero in terms of momentum but not in terms of energy. This is wasted energy.)
    There are other problems that sap available energy, but these two combined with the effective exhaust velocity [itex]v_{\text{eff}} = v_e + (P_e-P_a)A_e/q[/itex] pretty much explain why you get maximum thrust when (a) Pe = Pa and (b) Pa = 0.

    Naively, one would expect that maximizing exhaust pressure should maximize thrust. This ignores that adjusting the nozzle to reduce exhaust pressure can increase the exhaust velocity ve, particularly so when Pe > Pa.

    When Pe > Pa, the increase in thrust attributable to an increasing ve overwhelms the loss in thrust attributable to a decreasing Pe. The situation reverses when Pe < Pa. For a given ambient pressure, you get peak thrust when the exhaust pressure equals ambient. Decreasing ambient pressure means that you can get more useful kinetic energy out of a given fuel. You can get closer and closer to the ideal as ambient pressure decreases.

    Bottom line: For a given ambient pressure, maximum thrust is attained when exhaust pressure is equal to ambient. The ambient pressure that lets one get the very most out of a rocket is vacuum.
  5. Oct 26, 2011 #4


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    A couple of examples might help. One is the Apollo project, since there is plenty of film footage to demonstrate it. Each stage had a different nozzle design in order to work most efficiently in the atmospheric range that it was designed for. Huge bell-shapes on the Saturn 5, trumpets on the 2nd stage, and little pointy cones on the 3rd.
    One of my favourites, the linear aerospike engine, capitalizes upon the very properties of the atmosphere that it has to overcome.
  6. Oct 26, 2011 #5


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    The Saturn V (which is the whole vehicle, not just the lower stage) uses the same J-2 engines on the second and third stages. The second stage uses 5 of them, the third stage uses a single engine. In addition, the J-2 uses a high expansion bell nozzle (ε=28), as compared to the lower stage F-1 engine which has a lower expansion ratio (ε=16). Yes, the lower stage engines are physically larger, but that's because of the enormously higher mass flow rate, not the ambient conditions.
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