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boundary conditions for hub and shroud

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Old   March 23, 2009, 22:16
Default boundary conditions for hub and shroud
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Hey guys,
I am simulating a centrifual pump, but not so clear about the boundary conditions.
If it is a closed impeller, we should set hub and shroud as "relative velocity =0"? If open impeller, then hub and shroud are walls with "absolute velocity=0?
Is this understanding right? Thanks!
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Old   March 24, 2009, 01:24
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In a closed impeller, shroud will at rest while hub rotating about the axis. And there is no shroud in an open impeller. However, hub also spins round the axis. So it should be given a velocity value. Generally, we will specify a rotation speed for the ratating domain.
I am not sure I got your idea, but wish it helps.
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Old   March 24, 2009, 18:22
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Hi rikio,
Thanks for your reply.
For a closed impeller, shroud will "at rest"? Why is that? isn't it rotating together with the blade?
Thanks again for your explaination~
jasmine

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Originally Posted by rikio View Post
In a closed impeller, shroud will at rest while hub rotating about the axis. And there is no shroud in an open impeller. However, hub also spins round the axis. So it should be given a velocity value. Generally, we will specify a rotation speed for the ratating domain.
I am not sure I got your idea, but wish it helps.
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Old   March 24, 2009, 21:05
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Hi,

Only hub and blade rotate in a wheel-spin-only pump. Mean, wheel is the only moving part. It is easy to see if there is tip clearance since shroud and blade are separated.
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