If there was a standing sound wave tube and a flammable gas was introduced then ignited, would the combustion be more forceful and more efficient since its following a standing wave, than just a gas ignited within a tube?

  • $\begingroup$ What does "forceful" and "efficient" mean? $\endgroup$ – DumpsterDoofus Mar 18 '14 at 17:09
  • $\begingroup$ exertion of more force = forceful $\endgroup$ – user42744 Mar 18 '14 at 17:11
  • $\begingroup$ efficient - the gas is burning at a higher capacity due to the standing wave $\endgroup$ – user42744 Mar 18 '14 at 17:12
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    $\begingroup$ Although I appreciate accepting the answer, I recommend leaving it unaccepted for awhile. My answer isn't particularly great, just enough to get somebody started, and somebody may want to come along and give a really awesome answer. But they're less likely to do that if my answer is already accepted. So just wait a little while before clicking the check mark and you might get better answers! $\endgroup$ – tpg2114 Mar 18 '14 at 17:20

Although your terms are not precise, I have a feeling that what you are describing is a Rubens' Tube.

Yes, the combustion reflects the standing wave as shown below (from Wikipedia):

From Wikipedia

Recall that sound is made up of pressure waves and so a standing sound wave means there is larger pressure at some points and lower pressure at others. At the high pressure points, the fuel is pushed out through the opening faster resulting in a taller flame. Also, because the density is increased, more fuel comes out and combustion in enhanced. The opposite is true at the lower pressure regions.

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  • $\begingroup$ If the wave pattern were centered in the tube, using 3 equal tones set apart at 60 degrees, would this equalize the pressure at both ends of the tube assuming both ends were open. $\endgroup$ – user42744 Mar 18 '14 at 17:19

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