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It is generally said that, to obtain optimal specific impulse, a nozzle should be just the right size that the exhaust pressure is equal to the ambient pressure. Quoting this answer, a relevant equation is

$$F = q V_e + (P_e-P_a) A_e$$ Where $P_e$ is the exit plane pressure, $P_a$ is ambient pressure, and $A_e$ is the exit plane area. $qV_e$ without the correction term gives the thrust when the exit plane pressure matches ambient. Here $q$ is mass flow and $V_e$ is exit velocity.

Is it a valid interpretation to say that an overexpanded nozzle behaves like a suction cup? By this I mean that the thrust reduction from overexpansion is due to the ambient pressure on the outside of the nozzle being greater than the exhaust pressure in the inside of the nozzle (at least around the rim).

One thing I like about this explanation is that it neatly explains why even a sea-level-optimized engine gives better performance in a vacuum than at sea level. (The ambient pressure means that there is a smaller delta between the pressure inside the nozzle and the pressure outside it.) One thing I don't like about this explanation is that it suggests that a rocket engine's performance ought to change based on how fast the rocket is moving through the atmosphere (on account of dynamic pressure).

Obviously the mechanism that allows the pressure on the interior to be less than the ambient pressure is vastly more complex for a rocket engine than for a suction cup.

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  • $\begingroup$ Would you call the back end of a semi a "suction cup"? Or an airplane's spoilers? $\endgroup$
    – Darth Pseudonym
    Commented Nov 2, 2023 at 19:46
  • $\begingroup$ If you define (as you seem to) "acting like a suction cup" as "the thrust reduction from overexpansion is due to the ambient pressure on the outside of the nozzle being greater than the exhaust pressure in the inside of the nozzle" then yes by definition. That's what it says in the equation after all, you have just stated it in words. But suction cups usually don't have any thrust AFAIK so I am not sure how useful this definition is. $\endgroup$
    – Organic Marble
    Commented Nov 2, 2023 at 20:32
  • $\begingroup$ @OrganicMarble: A suction cup has net negative thrust (force), which is what holds it against a surface. If my interpretation is right, then a sufficiently overexpanded nozzle would also have negative thrust, which seems counterintuitive; at the least, there should be some theorem that flow separation is guaranteed for any nozzle big enough to produce negative thrust otherwise. $\endgroup$
    – Charles Staats
    Commented Nov 2, 2023 at 20:41
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    $\begingroup$ The pressure thrust term is typically dwarfed by the momentum thrust term. If an engine didn't produce enough momentum thrust to overcome the negative pressure thrust, it would be a pretty terrible engine. One could even say it sucked. $\endgroup$
    – Organic Marble
    Commented Nov 2, 2023 at 20:48
  • $\begingroup$ +1 for the pun, but I'm not sure I follow. My understanding is that the normal force between the nozzle and the exhaust gas is what creates pressure (on the nozzle interior) and momentum. The $qV_e$ term is needed because the "pressure term" does not account for how the pressure varies depending where you are in the nozzle. $\endgroup$
    – Charles Staats
    Commented Nov 2, 2023 at 21:04

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