Fluid Flow and Reynold's number

Question

I am confused about the conditions on the Reynolds number for an ideal fluid flow.

Now, an ideal fluid flow is approximated by real fluid flow with a high Reynolds number (except for if the flow converges at a small orifice, but ignore that). The Reynolds number being very large corresponds to the shear stress being negligible.

However, I also thought that ideal fluid flow is laminar, and that flow becomes turbulent for high Reynolds numbers. Turbulent flow is not characteristic of an ideal fluid flow, so there is an apparent contradiction, which I cannot resolve.

Where have I gone wrong?

Answer 1

The key is that ideal flows are only approximately equivalent to real flows at large Reynolds numbers. In fact, they are only really the "same" mathematically at an infinite Reynolds number, where viscous forces are not just negligible, but entirely non-existent.

In that limit, where there just are no viscous forces, there is also no turbulence. The energy cascade and dissipation rate in turbulent flows is due to the viscosity. If there is no viscosity, there is no turbulence.

So there is no real contradiction when you remember that ideal flow is only approximately equivalent to real flow for very high Reynolds numbers. The Reynolds number has to be so high, in fact, that turbulence doesn't have the physical mechanism it needs to form (viscosity).

These kinds of confusions crop up all the time when learning about ideal (or potential) flows. A related question and answer about wings generating lift in potential flow might help explain some additional ways we can make the ideal case behave like real cases.

Answer 2

My understanding is that an ideal fluid is one which has zero viscosity. But there is a big difference between an ideal fluid, and a real fluid at very high Reynolds number, even for very low viscosity. With zero viscosity, an ideal fluid can slip at the wall, and the axial pressure gradient will be zero. However, for a real fluid, irrespective of how low the viscosity gets, due to the no-slip boundary condition, the velocity at the wall is zero. This means that, for high Reynolds number, the velocity gradient and shear stress at the wall will become finite, and the fluid will experience a finite axial pressure gradient.

Regarding laminar flow, this occurs in real viscous fluids at relatively low Reynolds numbers, below that required for transition to turbulence. However, even in the laminar flow region, the fluid velocity profile in a pipe is not flat (as it would be for an ideal inviscid fluid). The no-slip boundary condition here too requires a finite velocity gradient at the wall, a finite shear stress at the wall, and a finite axial pressure gradient. So, in summary, the no-slip boundary condition in real fluids combined with a non-zero viscosity results in very different behavior for real fluid compared to ideal fluids. In boundary layer flows at high Reynolds number, the velocity profile in the region outside the boundary layer can approach that for inviscid flow, but within the boundary layer (the region that controls viscous drag), the behavior is quite different.

Answer 3

...I also thought that ideal fluid flow is laminar...Turbulent flow is not characteristic of an ideal fluid flow.

I am assuming that by ideal fluid you mean inviscid fluid. For a flow to be called turbulent, as distinct from a complicated laminar flow, it must possess a few characteristics (see A first course in turbulence by Tennekes and Lumley, chapter 1). Among others, it must contain a 3-dimensional distribution of vorticity, and it must be dissipative. In an ideal fluid, due to absence of viscosity, there is no vorticity generation at walls and other boundaries (because no-slip isn't obeyed). Therefore an ideal fluid of uniform density, which doesn't contain vorticity initially, will be free of vorticity for all subsequent times. Such a flow, no matter how complicated it might become due to inertial instabilities, cannot be called turbulent. If the ideal fluid had some distribution of vorticity to begin with, there is still no dissipation of mechanical energy of the flow due to absence of viscosity. Again as before, no matter how complicated the flow becomes, it can't be called turbulent.

Vorticity can be generated in an inviscid fluid by other means, for example due to unstable density differences. I am not aware of dissipative mechanisms other than due to viscosity. So perhaps absence of dissipation won't allow an ideal fluid flow to be called turbulent. But the ideal fluid flow may be as complicated as a turbulent flow.

I am not sure of @tpg2114 's claim that "...energy cascade and dissipation rate in turbulent flows is due to the viscosity." Dissipation I agree with, but energy cascade is due to inertial instabilities which (as far as I am aware) has no connection with viscosity. Viscosity dissipates mechanical energy of the flow, and sets the smallest scale at which turbulent fluctuations can occur. In a dissipative fluid, a continual supply of energy from outside is needed to maintain turbulence; in the absence of viscosity, such a continual supply isn't required (fluid's kinetic energy is conserved at all times), but energy cascade can still happen and the resulting flow may appear as complicated as a turbulent flow. However there isn't viscosity to set the smallest scale, so the velocity profile may develop singularities (since ideal fluids don't exist, it's all hypothetical anyway).

...an ideal fluid flow is approximated by real fluid flow with a high Reynolds number...The Reynolds number being very large corresponds to the shear stress being negligible.

That's correct. Viscous effects aren't negligible in certain portions of the flow (as @ChesterMiller pointed out), for example at rigid boundaries where a real fluid must obey no-slip condition. Away from such regions, the average flow may be approximated as that of an ideal fluid. Bernoulli equation is approximately valid for the average flow. (I presume you are familiar with Reynolds averaging of turbulent flows)