[samycfd]

Sure.

I was confused, please forget the two last posts I wrote nonsense..

[samycfd]

Hello,

I tried to discretize the problem using the approach you gave me.

To compute axd and adc in [(axd-axc)i+1/2 - (axd-axc)i-1/2]/dx, I need to discretize them - what I did - but I need some BC also ? because n.(ad-ac) is only for the final step ?

I am confused..

[FMDenaro]

No, just cancel out the terms in rhs and lhs of the equation when a section lies on a wall...You get a matrix with some lines modified according to neumann bc.s and the sourc e term modified congruently. That dose not imply You have fixed zero pressure normal derivative

[samycfd]

No that's not what I mean,

I have to compute that ( 1D )
[(dP/dx)i+1/2 - (dP/dx)i-1/2]/dx = [(axd-axc)i+1/2 - (axd-axc)i-1/2]/dx
with the BC : dp/dn = n.(ad-ac)

But first, I have to compute axd and axc. And for that on the edges I need some BC, no ? Do I just let 0 on the edges ?

[FMDenaro]

Yes, use not permeable and no-slip bc.s

[samycfd]

this is only valid for the lid driven cavity, how to deal with that for an other flow ?

[samycfd]

I guess it's not a solvable problem. but still I could assume that the value on the boundary is the same as the closest value. ( zero-gradient condition)

[FMDenaro]

well, you can consider any type of bc.s, (inflow, outflow, etc), some details are here

http://onlinelibrary.wiley.com/doi/1...d.598/abstract

[samycfd]

Thank you again for all your help.

I would like to know from where come the relation dP/dn = n.(ad-ac) ?
It's the Helmholtz Hodge decomposition ?

[FMDenaro]

Yes, correct

[samycfd]

I understood the definition of a projection operator which allows us to project any velocity field on his divergence free field. And then we apply this operator on Navier Stokes equation. But I don't understand the simplifications..

Here is the steady velocity field is w = - grad p - u(grad u) + mu lap u and then we compute P(w) ( P projection operator ) with hypothesis that div u = 0 ( divergence free) . But after I don't get it..to arrive to the pressure equation with BC

[FMDenaro]

Just apply the divergence operator to all the terms in the steady momentum equation. This writes

Div Grad P = Div f


The key is that the momentum equation can be seen as the Hodge decomposition

[samycfd]

Okay , so there is an analogy between :
divgrad p = div f
and
div grad q = div w

( if w = v + grad q , the field we want to project with div v = 0 ) .
But I don't understand for which mathematical reason we could get the same BC for p.

[FMDenaro]

Originally Posted by samycfd

Okay , so there is an analogy between :
divgrad p = div f
and
div grad q = div w

( if w = v + grad q , the field we want to project with div v = 0 ) .
But I don't understand for which mathematical reason we could get the same BC for p.

To have a unique (apart a constant) solution, it is mandatory that the Neumann bc.s are imposed according to the Hodge decomposition, that is projecting the decomposition along the normal direction to the boundary. The divergenze-free constraint automatically set int[S] n.v ds = 0

[samycfd]

I am confused. I read that http://www2.math.uni-wuppertal.de/~r...u/lecture6.pdf .
And on pages 9 & 10, to arrive to the same result they assumes no slip BC on all boundaries.
I was thinking it was working even without that?

[FMDenaro]

To be well posed the HHD problem, only one bc must be provided.
For the pressure problem this condition accords with the mass flow rate, therefore the normal condition, not the tangential (no slip) is required.
The tangential condition is used for the HHD problem expressed in terme of vorticity and stream function

[samycfd]

So I need to know all the normal BC for velocity to be able to write the equation for pressure with BC. so that's why in your paper you made different cases depending what kind of normal BC for velocity it is ..

[FMDenaro]

Originally Posted by samycfd

So I need to know all the normal BC for velocity to be able to write the equation for pressure with BC. so that's why in your paper you made different cases depending what kind of normal BC for velocity it is ..

I considered several possibilities, depending on the type of BC.s, you can or not ensuring a orthogonal decomposition.

[samycfd]

I don"t see to what the case 2 (n.a diff 0 ) leads for the pressure ?

[FMDenaro]

Originally Posted by samycfd

I don"t see to what the case 2 (n.a diff 0 ) leads for the pressure ?

in this case you still have a decomposition but you can no longer demonstrate the unicity and orthogonality of the decomposition