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I have trouble understanding why we classify an inviscid adiabatic incompressible flow along a streamline as isentropic

Thermodynamic definition $$dS = dQ/T$$ Adiabatic Invsicid $$dQ =0= dS$$

So no heat added or lost no change in entropy.

From Boltzman I am less clear how is it possible that we can have additional gradients within in the volume and more ordered momentum yet have no change in entropy

$$S=k_BlnW$$

If we looked at a two different control volumes enter image description here

Starting pressure $P_1$and final pressure$P_1$ are the same and equal flow rate Q in and out are the same but the presence of the venturi causes additional gradients and an increase in ordered kinetic energy from $v1$ to $v_2$ so it follows a reduction in available microstates $x,y,z,px,py,pz$ compared to an inviscid flow without a venturi.

Understanding that no flow is not in thermodynamic equilibrium how can we say that these flows have the same entropy ?

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In deriving this formula it was assumed that no energy is dissipated for viscous property and other related things.So since there is no energy dissipation and no energy providing agent is given we can safely say Q=0. As S=Q/t this implies S=0.So adiabatic process is now isentropic. And the increase in kinetic energy is at the cost of decrease in potential energy. So in your picture i think P2 is less than P1

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  • $\begingroup$ Here Q is change in energy and S is change in entropy. $\endgroup$ – kk9 Jan 14 '17 at 7:12
  • $\begingroup$ Thanks for your reply I get the thermodynamic explanation and it from energy conservation but I'm looking for an explanation from Boltzmann. there appears to be a discrepancy between them that I just don't understand . Bernoullis has caused a movement to a less probable state but this doesn't cause a change in Entropy? $\endgroup$ – Tom Chester Jan 14 '17 at 8:37

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