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-   -   Boundary Condition Problem (https://www.cfd-online.com/Forums/fluent/123366-boundary-condition-problem.html)

atk55442587 September 11, 2013 02:19

Boundary Condition Problem
 
Hi ANSYS users,

I am a new FLUENT user, this is my first post in CFD online, maybe this is a easy question.

there is a channel have a inlet and a outlet, the inlet velocity is a sine function depend on time, and the outlet just outflow.

As you know, the sine wave is positive at 0 to PI, negative at PI to 2*PI.
When time at PI to 2*PI, the velocity become the outlet velocity but reverse the direction, and the inlet become outflow.

Easy to say, the inlet and outlet boundary condition exchange.


I have a problem now. How to set the boundary condition as the example in fluent? Should I use UDFs?

BTW, I had tried using "Dynamic mesh -> event", but appear the error message "the BCs are same type", maybe it's a wrong way.

Please give me some suggest.

Thanks

AWalmsley11 September 11, 2013 08:51

With regards to the transient boundary condition, yes you will to use UDFs. There are quite a few examples around online for transient boundary conditions. Alas, none for what I need.

I can't help with the other issue, sorry I don't have experience with dynamic meshing as of this time.

quantities September 11, 2013 22:54

Maybe you can use velocity inlet and velocity outlet, or velocity inlet and pressure outlet.
Is the flow fluid compressible or not? The compressibility will affect this periodic flow process.

atk55442587 September 12, 2013 02:32

Thank quantities, AWalmsley11 reply.

Maybe I had found the way to solve this problem.

I let FLUENT calculate the time at 0 to PI, when time equal to PI, FLUENT calculate completed.

And I can change the boundary condition between inlet and outlet, and don't initialized solution, let FLUENT keep calculating time at PI to 2*PI.

This way can work, but I don't know it will cause what bad effect.

AWalmsley11 September 12, 2013 10:38

If it's done that way it will not have the sine function, it will be constant for that time. Think of it as it will appear as a digital signal rather than analogue in graphical terms.


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