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  3. boundary condition for hub and shroud

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boundary condition for hub and shroud

sriluta
sriluta over 2 years ago

I use Omnis 5.2 and the turbomachinery template.

I am studying an axial turbine with 15 blades without rotors.  The dominant flow phenomena occur in circumferential and axial directions rather than radial direction. Therefore the annulus radial heigh is neglected and the three-dimensional Simulation is unwrapped and reduced in a two- dimensional domain to reduce the computational cost. So if I use the symmetry boundary condition as hub and shroud walls, it means I consider no wall and then it is like a cascade.  So Applying a symmetry boundary condition means that the simulation assumes the physical conditions at these hub and shroud walls to be reflective or symmetric. In essence, it treats these walls as if they were not there, and any flow information that would be transferred through these walls is instead mirrored from the opposite side of the geometry.

"and then it is like a cascade": By treating the hub and shroud walls as symmetrical or reflective, the simulation effectively models a flow situation where the presence of these walls is not affecting the results. So In the simulation context I found for the boundary's shroud and hub among others the boundary condition mirror, would this be the correct boundary condition in my case ?

greetings :)

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  • domen
    +1 domen over 2 years ago

    Hi, you could use an Euler wall rather than a N-S one, basically a wall that allows the flow to slip along the boundary. 

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  • sriluta
    0 sriluta over 1 year ago in reply to domen

    thx

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  • sriluta
    0 sriluta over 1 year ago in reply to domen

    thx

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