# boundary conditions for hub and shroud

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 March 23, 2009, 22:16 boundary conditions for hub and shroud #1 Member   ^_^ Join Date: Mar 2009 Posts: 36 Rep Power: 10 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!

 March 24, 2009, 01:24 #2 Senior Member   Rikio Join Date: Mar 2009 Location: SH, China Posts: 182 Blog Entries: 1 Rep Power: 10 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.

March 24, 2009, 18:22
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Hi rikio,
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

Quote:
 Originally Posted by rikio 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.

 March 24, 2009, 21:05 #4 Senior Member   Rikio Join Date: Mar 2009 Location: SH, China Posts: 182 Blog Entries: 1 Rep Power: 10 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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