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I'm reading chapter 16 in Fluid Mechanics by Kundu.

It is stated (figures 16.16 and 16.17) that in a constant area duct flow with heating or friction, to go from subsonic conditions to supersonic violates the 2nd law of thermodynamics. How can that be possible? How does a fluid attain supersonic speeds in the first place? Somehow, one must be able to get ambient quiescent air to supersonic velocities (it's done all the time). I know that you cannot generate a normal shock by going from subsonic to supersonic, only the other way around, but why can't you speed a fluid from subsonic conditions to supersonic in the duct? In that case, there just would not be any shock generated? Would that violate the 2nd law?

I'm missing something here...

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Somehow, one must be able to get ambient quiescent air to supersonic velocities (it's done all the time).

Of course, and that same chapter already told you how it could be done: in convergent–divergent nozzle.

The statement

... the upper left branch of the solution $M_2 > 1$ when $M_1 < 1$ is inaccessible because it violates the second law of thermodynamics

applies only to constant area ducts with friction and heating.

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  • $\begingroup$ I just did not see the link between the normal shock solution and the constant area duct. What if there is friction in the variable area duct? Is it still a violation? If my converging diverging nozzle has friction? $\endgroup$ – l3win Oct 11 '13 at 5:19
  • $\begingroup$ Violation of what? 2-nd law of TD? No, of course. If the nozzle has friction it will results in entropy production, so the formulas for isentropic process would not apply. But if friction is small, then the effect will be minor and the resulting flow, though possibly slower than isentropic approximation could suggest would still be supersonic. $\endgroup$ – user23660 Oct 11 '13 at 6:52

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