Let us consider a 2D irrotational flow, such that $\nabla\times\boldsymbol u =\boldsymbol 0$. Defining the stream function such that $\boldsymbol u =\nabla\times\psi \boldsymbol n$ where $\boldsymbol n$ is the unit vector perpendicular to the plane of the motion.
The incompressibility condition implies $\nabla^2\psi =0$, so $\nabla^2 \boldsymbol u= \nabla^2 \nabla\times\psi \boldsymbol n=\nabla\times\nabla^2\psi \boldsymbol n=\boldsymbol 0$. From this follows that in the Navier-Stokes equations the term $\nu\nabla^2\boldsymbol u=0$ vanishes, no matter the viscosity.
Now, does this allow us to say that the fluid is inviscid?
Or, after considering the nature of the inertia, we can say that:
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if inertia can be neglected, the equation of the motion is $\nabla p=\boldsymbol 0$.
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if inertia is important, the flow is at high Reynolds number.
My problem is: Is every irrotational flow inviscid? This is kind of counter-intuitive. But I think my error is in the "no matter the viscosity"…
Best Answer
A version of your proof without a stream function:
The Laplace acting on the velocity may be expressed via the curl of the curl identity and aside from the $\nabla\times (\nabla\times \vec u)$ which vanishes, you also get another term $\nabla\cdot (\nabla\cdot \vec u)$ which vanishes (only) if one assumes incompressibility (it's the conservation of the mass).
So yes, one may neglect the viscous term if the flow is irrotational. In this sense, the irrational flows are automatically inviscid, too. I believe that this is the right solution to the problem 2 on page 143 of this Chapter
that is entirely dedicated to the inviscid and irrotational flows in this very combination. The converse isn't true. Inviscid flows may refuse to be irrotational: they may have vorticity.
However, I would still mention that the implication proved above isn't necessarily conceptually important. It's because while people sometimes discuss inviscid flows in which the whole viscous ($\mu \Delta \vec u$) term is zero or negligible, they're more likely to discuss "inviscid fluids". When we talk about the inviscid flows, we really want to claim that this term vanishes not because $\Delta\vec u=0$ but because $\mu$ is zero (or negligible). Even though the adjective "negligible" depends on the precise condition, typical speeds and Reynold's number etc., it's still meant to be a general property of a fluid rather than a property of some particular solutions. And of course, the fact that a fluid is able to exhibit irrotational flows does not mean that it is an inviscid fluid.