2
$\begingroup$

How is the exit velocity of the flow in a sounding rocket nozzle? Is it subsonic or supersonic?

Is a de Laval nozzle required for sounding rockets?

Bonus Question: What about Model Rockets?

$\endgroup$
  • $\begingroup$ Which sounding rocket? But, regardless of which one you are talking about, it is unlikely to be anything but supersonic. $\endgroup$ – Organic Marble Jan 29 at 19:38
  • $\begingroup$ Any jet-type engine needs a supersonic exhaust. Otherwise you're not using newtons 2nd, and the whole endeavour is a waste of money. $\endgroup$ – AtmosphericPrisonEscape Jan 29 at 22:49
  • $\begingroup$ @AtmosphericPrisonEscape Even subsonic nozzles use Newton's 2nd law, F = ma, as long as the flow velocity out of the nozzle is higher than the flow velocity into it, i.e. the flow accelerates. $\endgroup$ – Tom Spilker Jan 29 at 23:55
  • $\begingroup$ @TomSpiker But only supersonic flows accelerate upon expanding, so.... $\endgroup$ – AtmosphericPrisonEscape Jan 30 at 2:48
  • $\begingroup$ the vast majority of jet-type engines are subsonic @AtmosphericPrisonEscape $\endgroup$ – JCRM Jan 30 at 9:36
4
$\begingroup$

Essentially all practical rocket engines have supersonic exhaust velocities. This includes model rockets!

The force (thrust) generated by a rocket engine is given by $$F=\dot{m}V_e$$

where $F$ is the thrust, $\dot{m}$ is the rate of propellant mass flow from the nozzle (in units like kg/s), and $V_e$ is the exhaust velocity. Note that the force is directly proportional to the exhaust velocity. If you want to produce a certain level of thrust, doing it with $V_{e1}$ half as large as $V_{e2}$ means you have to feed twice as much propellant per second into engine 1 than for engine 2. Due to the exponential nature of the Tsiolkovsky rocket equation $$\Delta V=V_e \ln \frac{m_o}{m_f}$$ where $\Delta V$ is the velocity you need to impart to the rocket, $m_f$ is the mass of the rocket (and everything still attached to it) after the burn ends, and $m_o$ is the mass at the start of the burn ($m_o = m_f + m_p$, where $m_p$ is the mass of the propellant used), this means you have to carry more than twice as much propellant to get the same $\Delta V$, assuming $m_f$ doesn't change.

Or, if both $m_f$ and $m_p$ are fixed, the $\Delta V$ you'd get from engine 1 is half as much as you'd get from engine 2.

The net result: high $V_e$ is always good, unless getting higher $V_e$ adds significantly to $m_f$, as is sometimes the case with electric propulsion engines like ion engines. It is possible to build a subsonic nozzle for a rocket engine, but building a supersonic nozzle doesn't take much more mass (or money), so you wind up ahead with the supersonic nozzle, even for cheap little engines like model rocket engines.

There are many examples of subsonic nozzles in the world—but they aren't rocket engine nozzles.

$\endgroup$
  • $\begingroup$ Are there vocabulary terms for "supersonic nozzle" and "subsonic nozzle"? Probably something like "Rocket Nozzle" versus "Aeronautics Nozzle" but less layman? $\endgroup$ – Magic Octopus Urn Jan 31 at 20:23
  • 1
    $\begingroup$ @MagicOctopusUrn Not that I know of. Such nozzles, both subsonic and supersonic, have many varied applications outside of rocketry, so I don't think there's a nomenclature that would specify a supersonic nozzle as a rocket nozzle. That said, the usual design for a supersonic nozzle is the "de Laval nozzle". For examples of non-rocket applications of supersonic nozzles, they are used in supersonic wind tunnels, and even for condensing such gases as argon out of air. $\endgroup$ – Tom Spilker Jan 31 at 20:31
  • $\begingroup$ Typing "argon condensation", "supersonic wind-tunnel" and "de Laval nozzle" into wikipedia now! Thanks :) $\endgroup$ – Magic Octopus Urn Jan 31 at 20:32
  • $\begingroup$ @MagicOctopusUrn You'll find that the extraction of argon is a multi-step process. Liquifying air is done with the supersonic nozzle (though that's how they did it years ago—if there's a newer, more efficient process I don't know about it), then you have to "fractionate" it i.e. separate it into its component species. One way to do that is to slowly warm up the mixture. The different species have different boiling temperatures, so they come off one by one. $\endgroup$ – Tom Spilker Jan 31 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.