# Why is Bernoullis Isentropic

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

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 ?