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If one derives the Bernoulli Equation for the isentropic ideal gas flow you get:

$$ \frac{1}{2}v^2 + g z + \left(\frac{\kappa}{\kappa -1} \right) \frac{p}{\rho} = const. $$

Two questions:

  1. Is isentropic equivalent to frictionless?

  2. If the temperature stays the same during the process the quotient of pressure and density stays the same? This would lead to the very easy formula for the velocity at $t_2$. $$v_2=\sqrt{v_1^2-gz}$$

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  1. Yes, after a fashion, increase of entropy is always coupled to some kind of "friction". Though isentropic could be argued to be stronger than frictionless, as it does not only refer to mechanical friction. For example a Rayleigh flow with heat transfer will not be isentropic (credits to Azad for pointing out the example).

  2. Nearly, it is $v_2 = \sqrt{v_1^2 - 2g \Delta z}$. But note, that constant temperature does not necessarily imply isentropicity.

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  • $\begingroup$ @mcocdawc it's not equivalent. to be isentropic it must be frictionless (since friction is irreversible) but it's not sufficient. frictionless flow with heat transfer (rayleigh flow) is also non-isentropic. $\endgroup$ – Azad May 19 '15 at 18:59
  • $\begingroup$ The equation for v_2 looked "far too short". Especially since it is completely independent from any given cross section. Or is there an error in my interpretation. PS: Shall I edit the wrong formula? $\endgroup$ – mcocdawc May 19 '15 at 18:59
  • $\begingroup$ @Azad thanks, I worked this into the answer. Non-ballistic heat transfer is of course also driven against a "friction" which causes the entropy to increase. $\endgroup$ – Sebastian Riese May 19 '15 at 20:28

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