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March 28, 2014, 20:04 |
slip boundary's equation.
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#1 |
Senior Member
Dongyue Li
Join Date: Jun 2012
Location: Beijing, China
Posts: 838
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Hi guys,
In inviscid flow, wall's boundary is slip. Lets take account of velocity: the equation of velocity in the slip boundary should be: The reason is that: n_x=0.n_y=-1,u_y=0. But in user guide it said: Code:
The normal vector is fixedValue=0, the tangential vector is zeroGradient; So how can I derive that the tangential vector is zeroGradient? Thanks |
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March 28, 2014, 22:03 |
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#2 |
Member
xuhe-openfoam
Join Date: Aug 2013
Location: DaLian,china
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hello
for incompressible fluid: i think we could deduce "the tangential vector is zeroGradient" from equation of continuity . Best Regards xuhe |
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March 29, 2014, 20:28 |
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#3 |
Senior Member
Dongyue Li
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March 30, 2014, 01:46 |
may this help you !
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#4 | |
Member
xuhe-openfoam
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Quote:
is it clear ? |
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March 30, 2014, 03:05 |
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#5 |
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xuhe-openfoam
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do you deal with twophase fluid ?
xuhe |
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March 30, 2014, 06:23 |
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#6 |
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Dongyue Li
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March 31, 2014, 03:21 |
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#7 |
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Bernhard
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Your equations are not correct. v(y=0)=0 does mean dv/dy=0 per se (i.e. if v(y=1)=1). Therefore, your statement that du/dx=0 is not correct. A slip boundary is impermable (thus v=0) and stress-free (thus du/dy=0, for your sketch).
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March 31, 2014, 03:47 |
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#8 |
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xuhe-openfoam
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March 31, 2014, 03:50 |
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#9 | |
Senior Member
Dongyue Li
Join Date: Jun 2012
Location: Beijing, China
Posts: 838
Rep Power: 17 |
Quote:
Just because its stress free. so u(y=0) has no impact on u(y=1). So this du/dy=0. It makes sense. Thanks. |
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