[samycfd]

Hi everyone,

What's the best solution to get the pressure field from the velocity field ?
Solving the pressure poisson equation (PPE) ?

( I will assume the boundary for conditions for pressure depending on each flows. )

Thank you very much !

samuel

[FMDenaro]

Originally Posted by samycfd

Hi everyone,

What's the best solution to get the pressure field from the velocity field ?
Solving the pressure poisson equation (PPE) ?

( I will assume the boundary for conditions for pressure depending on each flows. )

Thank you very much !

samuel

is the velocity field divergence-free? If yes, you can solve the Poisson equation, however be careful your pressure will be defined apart a function of time

[samycfd]

The velocity is assumed to be divergence-free ( the gradient of the out-of-plane component is very small ) .

I have solved the PPE but still get some wrong results compare to a fluent simulation. I am studying only steady cases so the time dependence does not matter.

I am using finite differences to compute the RHS of the PPE ( Lap(P) = ux^2 + uxvy + uy^2 ). Are there any issues there ? ( I am using forward differences ).

[FMDenaro]

do you know the 3 velocity components in the same nodes of a Cartesian box? Are they equidistant?
First of all, check the (Div v) value in each node using central second order formula. Then, consider the steady momentum equation:

Grad (p/rho0)= Div (2ni*Grad v - vv)and take the divergence of both sides to write the Poissone equation for pressure.
Be carefull that boundary conditions must be properly posed on the frontier

[samycfd]

The problem is that I not always know the three components. But if just know two of them, the case could often be assimilated to a planar case( in the flows I am going to study.. of course it's not always right!)

I have checked the divergence of the flow.

What represent ni and vv in your equation ?

If I take the divergence of both sides of the navier stokes equation

grad(P) = 1/rho * [ - grad(v) v + nu * lap(v)]

and assuming div(v) = 0 , then at the second order I get

lap(P) = - 1/rho * ( u_x^2 + u_x * v_y + v_y ^2 )

( where v=(u,v) and u_x represent the derivative along the x axis of u and so on .. )

Is that right ?

[FMDenaro]

Originally Posted by samycfd

The problem is that I not always know the three components. But if just know two of them, the case could often be assimilated to a planar case( in the flows I am going to study.. of course it's not always right!)

I have checked the divergence of the flow.

What represent ni and vv in your equation ?

If I take the divergence of both sides of the navier stokes equation

grad(P) = 1/rho * [ - grad(v) v + nu * lap(v)]

and assuming div(v) = 0 , then at the second order I get

lap(P) = - 1/rho * ( u_x^2 + u_x * v_y + v_y ^2 )

( where v=(u,v) and u_x represent the derivative along the x axis of u and so on .. )

Is that right ?

but if you don't have the complete 3D velocity field the problem is big...
However, you have to discretize the equation as

Div (Grad P) = 1/rho * Div [ Div ( 2*nu *Grad v - vv)]

vv is the convective flux. You cannot assume div v =0 without ensuring that is numerically satisfied in each node

[samycfd]

Yes of course but I am asked to qualify in which extent this assumption ( divergence-free) is still right by maybe defining a value for the divergence above which results will be wrong. But anyway I will deal with that and I have understood the problem.


So I will try this equation !
But I don't understand why I have to discretize the equation in this way ? Is the problem about numerical instabilities or convergence of the equation ?


Thank you very much for your answers !

[FMDenaro]

Originally Posted by samycfd

Yes of course but I am asked to qualify in which extent this assumption ( divergence-free) is still right by maybe defining a value for the divergence above which results will be wrong. But anyway I will deal with that and I have understood the problem.


So I will try this equation !
But I don't understand why I have to discretize the equation in this way ? Is the problem about numerical instabilities or convergence of the equation ?


Thank you very much for your answers !

If you integrate that equation over the surfaces of the domains, provided that the BC.s are correct, you satisfy the compatibility relation that ensures the existance of a solution.
If you want, more details are in http://onlinelibrary.wiley.com/doi/1...d.598/abstract

[samycfd]

Do you know where I can find the full-text online ? Or could you send me the paper by private message ( because I ve figured out that you were the author ) ?

Thanks !

[FMDenaro]

https://www.researchgate.net/publica...ns?ev=srch_pub

[samycfd]

Thank you, I will take a look at it and try to understand all of that !

Best regards
Samuel

[samycfd]

You would mean Div (Grad P) = 1/rho * Div [ Grad( 2*nu *Grad v - vv)] ?because we could not take the divergence of a scalar value.

Do you agree with the fact that the convective term is ( v . grad ) v ?

Thanks

[FMDenaro]

Originally Posted by samycfd

You would mean Div (Grad P) = 1/rho * Div [ Grad( 2*nu *Grad v - vv)] ?because we could not take the divergence of a scalar value.

Do you agree with the fact that the convective term is ( v . grad ) v ?

Thanks

no, from the Reynolds theorem you write Div (vv) that is a vector (therefore you still can calculate divergence).... only if you use the mass equation then you can rewrite the momentum equation in quasi-linear form

[samycfd]

I don't understand how is it possible, because the divergence returns a scalar ...
I don't get your point .

Before using the mass equation, after taking the divergence you normally get something like : div ( grad P ) = div ( nu lap(v) - vv )

[FMDenaro]

Originally Posted by samycfd

I don't understand how is it possible, because the divergence returns a scalar ...
I don't get your point .

Before using the mass equation, after taking the divergence you normally get something like : div ( grad P ) = div ( nu lap(v) - vv )

no, vv is a tensor (as Grad v), therefore its divergence is a vector

The diffusive flux is 2*nu*Grad v, when you take the divergence it writes nu*Lap v only if Div v is fulfilled.
Therefore

( grad P ) = Div ( nu 2*nu*Grad v - vv )

-> Div ( grad P ) = Div [Div ( nu 2*nu*Grad v - vv )]

[samycfd]

I have understood one part but why the convective term is a tensor ?

Grad v is a tensor, in 2D we get : Grad v = [ ux uy ; vx vy ] ( Matlab notation )

but the convective term defined as ( v . grad ) v gives a vector for me :

( v . grad ) v = [ u ux + v uy ; u vx + v vy ]

I must be wrong somewhere but I don't know why

[FMDenaro]

Originally Posted by samycfd

I have understood one part but why the convective term is a tensor ?

Grad v is a tensor, in 2D we get : Grad v = [ ux uy ; vx vy ] ( Matlab notation )

but the convective term defined as ( v . grad ) v gives a vector for me :

( v . grad ) v = [ u ux + v uy ; u vx + v vy ]

I must be wrong somewhere but I don't know why

the convective flux of the momentum quantity is a tensor as it represents the mass for volume unit (rho*v) macroscopically transported by the velocity field v.
I suggest have a reading of some basical fluid mechanics textbook

[samycfd]

So you're talking of something different from the convective term in Navier-Stokes equation ?

Because for me it's that ( 3.1 section )
http://en.wikipedia.org/wiki/Navier%...e_acceleration

[FMDenaro]

Originally Posted by samycfd

So you're talking of something different from the convective term in Navier-Stokes equation ?

Because for me it's that ( 3.1 section )
http://en.wikipedia.org/wiki/Navier%...e_acceleration

the difference is just in the conservative or quasi-linear form of the NS equations, however both v*Grad v and Div (vv) produce a vector.
For the formulation of a general transport equation, the Reynolds transport theorem is the corner stone. From that you can derive everything...
Have a reading of some good textbook, wikepedia is very poor...

[samycfd]

I know it was just because there formula were written clearly.

In a lot of great books I found the Pressure poisson equation in 2D for a divergence-free flow as the following : ( approx ~ 2nd order )

laplacien(p) = - rho * ( (u_x)^2 + u_x * v_y + (v_y)^2 )
+ Boundary conditions

I want to be certain, you told me that it is not correct ?