# Boundary condition: intermediate velocity for fractional time step method

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 June 19, 2015, 18:37 Boundary condition: intermediate velocity for fractional time step method #1 New Member   Rime Join Date: Jun 2015 Posts: 26 Rep Power: 2 Hi! How to calculate u*=u_n+1 for the intermediate velocity u*

June 20, 2015, 03:47
#2
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Filippo Maria Denaro
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Quote:
 Originally Posted by Rime Hi! How to calculate u*=u_n+1 for the intermediate velocity u*

are you talking about the intermediate BC.s in the fractional time step method? The literature is full of paper about this issue, starting from the hystorical paper of Kim & Moin on JCP you can find many many studies and proposals.

However, u_n+1 is the physical BC you have to know, a zero-order approximation simply set it as u*.

 June 20, 2015, 07:25 #3 New Member   Rime Join Date: Jun 2015 Posts: 26 Rep Power: 2 Thank u for the reply. yes for fractional time step method

 June 20, 2015, 14:24 #4 New Member   Rime Join Date: Jun 2015 Posts: 26 Rep Power: 2 I use FSM with free pressure boundary : 1)n.u*=n.u_n+1 u*=u_n+1 => u*=u_n+dt*u_t =u_n+dt*(-u*grad_u+(1/rho)*(mu*laplace(u)+grad_pression+F) is correct?? (depent to density rho) 2) t.u*=t.(u_n+1+(dt*grad_Phi) 3) for pressure: laplace(phi)=(rho/dt)*div u^* boundary n.grad_phi=0 4) and to calculate u_n+1 u_n+1=u*-(dt/rho)*grad_phi No boundary conditions, ie I compute the velocity for interior and boundary points. PS: I am beginner in CFD I need your help Thank you

 June 20, 2015, 18:27 #5 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 1,854 Rep Power: 25 I think your process need a careful revision... 1) Are you using a second order implicit time integration in the momentum equation? 2) The pressure equation is nothing else but the continuity equation in which the Hodge decomposition is substituded. 3) You need to take care of the product Div Grad, I suggest not to write it as Laplacian for a correct setting of the Neuman BC.s 4) What books/papers are you studying for the FTS method?

 June 20, 2015, 18:48 #6 New Member   Rime Join Date: Jun 2015 Posts: 26 Rep Power: 2 for time discretisation I use Adams Bashforth and Crank Nicolson scheme. [1] Application of a Fractional-Step Method to Incompressible Navier-Stokes Equations J. KIM AND P. MOIN [2] David L. Brown, Ricardo Cortez,y and Michael L. Minionz Accurate Projection Methods for the Incompressible Navier–Stokes Equations [3]J.L. Guermond, P. Minev c, Jie Shen An overview of projection methods for incompressible flows and other paper for spatial discretisation (Radial basis function).

June 21, 2015, 03:21
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Filippo Maria Denaro
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Quote:
 Originally Posted by Rime for time discretisation I use Adams Bashforth and Crank Nicolson scheme. [1] Application of a Fractional-Step Method to Incompressible Navier-Stokes Equations J. KIM AND P. MOIN [2] David L. Brown, Ricardo Cortez,y and Michael L. Minionz Accurate Projection Methods for the Incompressible Navier–Stokes Equations [3]J.L. Guermond, P. Minev c, Jie Shen An overview of projection methods for incompressible flows and other paper for spatial discretisation (Radial basis function).

Ok, therefore you solve the implicit scheme for v* using the BC.s v* = v_n+1 + dt Grad phi_n, right?

Then you solve the elliptic equation Dic Grad phi_n+1 = Div q, wherein q is the source term.

 June 21, 2015, 07:01 #8 New Member   Rime Join Date: Jun 2015 Posts: 26 Rep Power: 2 Thank you for the reply. my result diverge and when I use steady boundary I get a false solution.

June 21, 2015, 07:10
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Filippo Maria Denaro
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Quote:
 Originally Posted by Rime Thank you for the reply. my result diverge and when I use steady boundary I get a false solution.

check if you satisfy the continuity equation in each node just after the first time step.

 June 21, 2015, 07:16 #10 New Member   Rime Join Date: Jun 2015 Posts: 26 Rep Power: 2 I get div u around 0.04

June 21, 2015, 07:29
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Filippo Maria Denaro
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Quote:
 Originally Posted by Rime I get div u around 0.04

you have some bug in the code, I think in setting the BC.s for the pressure equation.
Are you setting:

d phi/dn = (1/dt) ( v*-v_n+1).n

in the Poisson equation?

June 21, 2015, 07:42
#12
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Rime
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Quote:

June 21, 2015, 08:31
#13
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Filippo Maria Denaro
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Quote:
 Originally Posted by Rime So I use n.grad_phi=0
have you modified the source term in a consequent way?

June 21, 2015, 08:36
#14
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Rime
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Quote:
 Originally Posted by FMDenaro have you modified the source term in a consequent way?
sorry, I don't understand the question

June 21, 2015, 08:43
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Filippo Maria Denaro
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Quote:
 Originally Posted by Rime sorry, I don't understand the question
the source term in the Poisson equation has been modified in a way congruent to the setting d phi/dn=0? How do you compute it?

June 21, 2015, 09:19
#16
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Rime
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 Originally Posted by FMDenaro the source term in the Poisson equation has been modified in a way congruent to the setting d phi/dn=0? How do you compute it?
so

I look that in my code I consider that the velocity in the boundary equal to zero (therefore that is false)

June 21, 2015, 11:58
#17
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Filippo Maria Denaro
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Quote:
 Originally Posted by Rime div grad_phi = Div u^* n.u^* = n.u_n+1 + n.grad_phi so n.grad_phi= n.(u^* - u_n+1^*) I look that in my code I consider that the velocity in the boundary equal to zero (therefore that is false)

As you see, if you set in n.u^* = n.u_n+1 + n.grad_phi the BC.s :

you MUST set also

n.u^* = n.u_n+1

therfore in computing the source term

Div u^*

you have to use the physical BC.s n.u_n+1.
One you have fulfilled such constraints, you have the compatibility condition satisfied and the Poisson equation will converge.
Finally, if you use a staggered secondo order discretization you should have the continuity equation satisfied up to machine precision

June 21, 2015, 12:34
#18
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Rime
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 Originally Posted by FMDenaro n.u^* = n.u_n+1 therfore in computing the source term Div u^* you have to use the physical BC.s n.u_n+1.
Thank you for your time, very useful.

n.u^* = n.u_n+1 <=> n.u^* = n.u_bc (bc=boundary condition)
because I can't see the difference between u_n+1 and steady u_bc

June 21, 2015, 12:37
#19
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Filippo Maria Denaro
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Quote:
 Originally Posted by Rime Thank you for your time, very useful. n.u^* = n.u_n+1 <=> n.u^* = n.u_bc (bc=boundary condition) because I can't see the difference between u_n+1 and steady u_bc

well, that's ok if the BC.s are steady. It must work

June 21, 2015, 12:44
#20
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Rime
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Quote:
 Originally Posted by FMDenaro well, that's ok if the BC.s are steady. It must work
Thank you, I will try again.
I hope it works

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