Guys I think I have a lame question about fluent.

When you have to choose a viscous model, you have a choice between inviscid, laminar or any turbulent model. Does it mean that an inviscid flow is never considered by fluent as turbulent? Why that? To me ONE example of an inviscid flow is a flow with a very high reynolds number (I know it s not the definition, but it s a good approximation). But a high reynolds number means that the flow can not be considered as laminar. At high reynolds number the little fluctuations in the BC are growing exponentially and cause turbulence effect downstream. The inviscid model does not include those little fluctuations.

Inviscid can't be turbulent... turbulence a viscous effect and inviscid means it has no viscosity... you talk about a high reynolds number, but reynolds number is rho*V*L/mu... if mu is 0, then Re would be infinite... I'm assuming when you talk about "flow with a very high reynolds number" you're talking about high speed flows. Some people make the assumption of inviscid flows for high speed external aerodynamics, but that's because the pressure forces are much stronger than the viscous forces on the body. They're incorporating some level of error, but for a first assumption it might be reasonable (depending on the geometry of course).

I hope this helps, Jason

I might be wrong, but I thought turbulence is due to little fluctuations growing exponentially (if Reynolds number is large enough) due to nature of Navier Stokes equation. So the origin of turbulence lays in the flow and the fluctuations of BC and not in the fluid properties (even if reynolds is linked to the fluid property). For me, if we take an inviscid fluid, viscosity is close to 0 and the Reynolds number is very large (tends toward infinity). Hence the Reynolds number is big enough to enable the tiny fluctuations to grow exponentially and cause vorticity and come up with a turbulent flow. So my question is why Fluent overlooks the fact that inviscid flow should be turbulent? My assumption is that the inertial forces are so large (due to high velocity) that any eddies are negligible compared to what the bulk flow is. But I am really not sure of it. Can someone confirm or bring up another idea or tell me that I am saying crap and I should better open a fluid mechanic book before bothering you guys (even though I promise I did open a fluid mechanic book)

Thank you

Ofcourse your assumption is correct. There are no eddies considered in an inviscid flow. In a real fluid motion, the turbulent flow is characterised by eddies(mixing or irregular fluctuation) which are super imposed on the main stream. These fluctutaions in the mean flow will cause additional stresses called "Reynolds Stresses" which will introduce an additional viscous effect termed as "molecular momentum viscosity" The effect caused by this is the apparent increment in the viscosity of the flow as compared to the laminar flow. At large Reynolds numbers, as you assumed, there exists a continuous transport of energy from the main flow into the large eddies. However, the energy is dissipated predominantly by the small eddies and the process occurs in the boundary layer where viscous laminar-sublayer exists. While in the inviscid flow, the theoretical assumption states that all the shear stress are considered negligible and the normal stress corresponds to the pressure acting on the fluid element. Hence in inviscid calculatioins the effect of the shear stresses in the laminar flow (due to the viscosity of the fluid)as well as the stresses in the turbulent flow (due to viscocity + Reynolds stresses, where, latter>>former) are not taken into account. Hence you cannot assume any eddies to be formed in an inviscid flow..and one important thing is INVISCID FLOW IS A SIMPLIFICATION OF THE ACTUAL FLOW and hence in real fluid motion, eddies do form in a high speed flow. So if softwares like Fluent or any analytical calcuations consider the flow as inviscid for the purpose of simplifying the calculations, there is no question arises whether they have to consider the flow to be laminar or turbulent as the effects of which on the flow are mainly due to the shear and Reynols stresses. Hence as you told there are no "exponentilaly" growing fluctutaions in an imaginary "Turbulent" flow in an inviscid flow. I used imaginary because you cannot categorize an inviscid flow as Turbulent or Laminar refering to the velocity of the flow.

Thank you very much for your enlightning answer. If I summarize:

turbulence form for large Reynolds number

for extremly high reynolds number, the eddies can be neglected, so the flow can be considered as inviscid so the flow doesn't need to be treated as turbulent as a first approximation.

So it means that turbulence needs to be considered for

very large number > Reynolds > large number

Right? Thank you anyway guys for all your contributions

This is what Jason has mentioned. When you are assuming the shear stresses to be zero (inviscid flow), you are neglecting the effect of viscosity (mu=0), that means you end up with infinite Reynolds Number. This doesnot mean that the flow is of very high speed. It only says that the pressure forces (inetria)are infinitely large compared to viscous forces. That may be true even for a low speed flow of a gas with very low viscosity. But in general, people use this(inviscid assumption) for very high speed flows with some probable error in the calculation, as Jason pointed out. This concept of inviscid flow seems to be little bit confusing because of the non-physical nature of the assumption of no shear stress in the flow. This is just for the simplification of the Navier-Stokes equation where if you neglect the viscous terms, you will be led to the Euler's equation which describes the inviscid flow. Obviously, the purpose of using Euler's equation is to reduce the order of N-S equations as a first approximation to the latter. One more ineteresting thing! It is not new that people are confused with the assumption of no viscosity! You migh't have heard about the well known D'Alembert's paradox, which says that there will be no drag force on a symmetrical airfoil at zero degree angle of attack. This is a non-physical condition arised due to neglecting the viscosity and hence the circulation around the airfoil, which is resposnsible for lift induced drag in actual case.

Yes I heard aout this paradox. Thank you Jason and Rajesh for your enlightments. Sylvain