I've been looking into rocket propulsion a bit and got stumped on something. I read that the thrust generated by an engine can be determined by the following: $$ F = \frac{w}{g}v_{e}+A_{e}(P_{e}-P_{a}) $$ where $\frac{w}{g}$ is the mass flow rate, $v_{e}$ is the exhaust speed, $A_{e}$ is the exit area, $P_{e}$ is the exit pressure and $P_{a}$ is the ambient pressure. Apparently the max thrust can be obtained when the exit and ambient pressures are equal as this makes the second term zero. I don't understand this. Wouldn't you want to maximize $P_{e}$ so as to increase the value of $F$?

Hope this isn't a dumb question...


1 Answer 1


The hard part is that $P_e$ isn't a completely independent variable. As the gas expands past the throat, thermal energy is being converted into kinetic energy. The gas cools down and speeds up.

So if you shorten the nozzle (creating an underexpanded flow), there is greater pressure at the exit (good). But the exhaust speed $v_e$ is lower (bad). The $\frac wg v_e$ term will be smaller and total force will be smaller.

  • $\begingroup$ If I'm not mistaken, a (properly shaped) longer nozzle on a vacuum engine always increases thrust because more expansion is contained and directed. Complicating factors such as thrust-to-weight ratio probably have to be factored in to constrain engine design in this way. $\endgroup$ Commented May 31, 2020 at 23:19
  • $\begingroup$ It's worth mentioning that the anticorrelation between pressure and velocity is a general principle – though it plays out quite differently in case of the supersonic flow in the expanding part of the de Laval nozzle, than in the more familiar “garden hose nozzle” scenario. $\endgroup$ Commented Jun 1, 2020 at 0:33

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