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slip boundary's equation.

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Old   March 28, 2014, 21:04
Default slip boundary's equation.
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Dongyue Li
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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:

u*n=0


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;
I can understand the first sentence. Because u_y=0.

So how can I derive that the tangential vector is zeroGradient?

Thanks
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Old   March 28, 2014, 23:03
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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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Old   March 29, 2014, 21:28
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Quote:
Originally Posted by bieshuxuhe View Post
hello
for incompressible fluid:
i think we could deduce "the tangential vector is zeroGradient" from equation of continuity .

Best Regards
xuhe
Could you elaborate it more? I want to know how can I get
\nabla u_x = 0
Thanks.
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Old   March 30, 2014, 03:46
Default may this help you !
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Quote:
Originally Posted by sharonyue View Post
Could you elaborate it more? I want to know how can I get
\nabla u_x = 0
Thanks.
look down the picture !
is it clear ?
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Old   March 30, 2014, 04:05
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do you deal with twophase fluid ?



xuhe
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Old   March 30, 2014, 07:23
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Quote:
Originally Posted by bieshuxuhe View Post
do you deal with twophase fluid ?



xuhe
cool, wonderful.

Yep, Im using solver on two-fluid model.

Thanks.
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Old   March 31, 2014, 04:21
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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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Old   March 31, 2014, 04:47
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Quote:
Originally Posted by Bernhard View Post
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).
thank you very much!

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Old   March 31, 2014, 04:50
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Quote:
Originally Posted by Bernhard View Post
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).
Oops, We should be more care about it. Yep, it should be du/dy=0 not du/dx=0.

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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