Boundary Condition Problem

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 September 11, 2013, 01:19 Boundary Condition Problem #1 New Member   Shu Shu Chang Join Date: Sep 2013 Location: Taiwan, Tainan Posts: 2 Rep Power: 0 Sponsored Links 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

 September 11, 2013, 07:51 #2 New Member   Andrew Walmsley Join Date: Sep 2013 Posts: 5 Rep Power: 5 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.

 September 11, 2013, 21:54 #3 New Member   Join Date: Sep 2013 Location: Shanghai China Posts: 6 Rep Power: 5 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.

 September 12, 2013, 01:32 #4 New Member   Shu Shu Chang Join Date: Sep 2013 Location: Taiwan, Tainan Posts: 2 Rep Power: 0 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.

 September 12, 2013, 09:38 #5 New Member   Andrew Walmsley Join Date: Sep 2013 Posts: 5 Rep Power: 5 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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