I was wondering. How can we set the flow as the inviscid flow? When we define "Fluid Domain", in "Fluid Editor", we will find 'Properties". If I set the "Dynamic Viscosity" in this "Properties" form to zero, are all behaviors of this flow inviscid? Thank you very much.

Atit

Atit,

Yep, that's usually the way, but don't use zero that could be nasty. Just use something really small, say 1.0E-10 or less.

Neale

If we just set the boundary condition at the wall as 'No-slip". Is that enough? Thank you very much.

Atit

I do think that it is necessary to do both thing, else it is meaningless )) Jan

Sorry it is the typo. I have to type 'Free Slip', not 'No-Slip.

Atit

It should be enough to set your walls to free slip and select Laminar flow. Don't define your viscosity to be very small or you will have severe convergence problems. Leave the viscosity of the fluid as it should be. As your solution converges, viscous effects will disappear.

I have done this to compare to inviscid airfoil calculations and it works well. Just keep in mind, however, that this is still a viscous code.

Regards, Robin

Thank you very much. Your suggestion is invaluable.

Atit.

I do not completely agree with you Robin, I have used this approach for many years ago, so my memory might be a little rusty, but I do believe that the viscosity should be set to a small value such that the viscous forces are eliminated. A robust solver should not have any problems with that ))). Regards Jan

The problem is not with solver robustness. If you specify a very small viscosity, you will encounter round-off problems when dividing by such a small number. Besides, the viscosity will help converge the solution faster, if anything you should increase it. If you boundary conditions do not introduce any rotational flow, the final solution will be inviscid anyway.

If you still do not agree, try it.

Robin

))))) I am thinking, if I dare Happy New Year What if the geometry create a rotation of the flow! Jan

You should define the viscosity to be a small number regardless of the boundary conditions. There are other mechanisms for introducing vorticity besides the boundary conditions. For example, a curved shock wave will introduce vorticity.

Neale

Response to both Jan and Neale:

Granted that shock waves and backward facing steps, etc. will introduce viscous effects. But if the flow is not irrotational (inviscid) due to boundary conditions, then you cannot and should not be interested in solving it as an inviscid flow! Sure, if you make your viscosity small you will get the same wrong answer for these as you would expect from and inviscid code!

The problems that arise in the solver due to your small viscosity are purely numerical and have nothing to do with robustness. If you assume there are no viscous effects, there are simplifications which may be made for solving inviscid flows. I don't need to remind you that you are dealing with a viscous flow code, so these optimizations are not done.

All I am suggesting is that if you are setting up a simple inviscid flow problem and want to solve it as such, then you can do so simply by applying the appropriate boundary conditions. At the wall, this means using a free-slip boundary condition.

Quite frankly, this is only of academic interest. Why would you want an inviscid flow solution at the cost of solving a viscous flow?

Enough said.

Happy New Year!

Robin

Dear Robin,

I am the newcomer in CFD. I hope that you have a time to give me the explanations of my new questions, occurred from your previous answers. They are:

- You said that if we set viscosity be very small, we will get severe convergence problem. Why sir? I guess that CFX may use the Reynolds number in computation. The Reynolds number approaches infinity when we set viscosity equal zero.

- How can 'shock waves' and 'backward facing steps' have the viscous effects? Are they only the effects in numerical approach or they also have the effects in theoretical approach? What are those effects?

- You said that 'if the flow is not irrotational (inviscid)'. That make me feel like 'irrotational' and 'inviscid' is the same word. But they are not, are they?

- From your latest answer, you say that

'Granted that shock waves and backward facing steps, etc. will introduce viscous effects. But if the flow is not irrotational (inviscid) due to boundary conditions, then you cannot and should not be interested in solving it as an inviscid flow! Sure, if you make your viscosity small you will get the same wrong answer for these as you would expect from and inviscid code! '

I am sorry but I do not understand all sentences except the first sentence. Could you please give me more explanation?

- Because of my very bad English, I do not understand your last comment.

'Why would you want an inviscid flow solution at the cost of solving a viscous flow?'

Did you ask me that why don't I write the inviscid solver to solve my problem? I think that my understanding is wrong. However if that is the real meaning of your question. My answer is 'this is the first step of my computation'. After I get the solution, I will increase the complexity of the problem grudually. Finally this problem will be the viscous flow.

I appreciate your kind and informative previous answers very much.

Atit