# Minimum Time Step with Navier Stokes

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 November 29, 2013, 23:48 Minimum Time Step with Navier Stokes #1 New Member   Join Date: Nov 2013 Posts: 5 Rep Power: 12 Hello, I am a bit of a newbie at CFD. I am trying to determine what the minimum time step would be when solving the N-S equations for a lid-driven cavity flow (2D) in order to achieve stability. The flow is incompressible, viscous, and the results I am interested in are at steady state. I am using a finite difference method with a co-located grid. I'm really not sure how to even approach this problem and any advice on what I need to do to find the time step would be greatly appreciated. Thank you very much for any and all replies.

November 30, 2013, 04:27
#2
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Filippo Maria Denaro
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Quote:
 Originally Posted by Atreides Hello, I am a bit of a newbie at CFD. I am trying to determine what the minimum time step would be when solving the N-S equations for a lid-driven cavity flow (2D) in order to achieve stability. The flow is incompressible, viscous, and the results I am interested in are at steady state. I am using a finite difference method with a co-located grid. I'm really not sure how to even approach this problem and any advice on what I need to do to find the time step would be greatly appreciated. Thank you very much for any and all replies.
The numerical stability region (Reh,cfl) is determined by the type of numerical scheme you use, each one producing a different map of stability.
I don't see your type of discretization, Re number and number of cells.
I can suggest a rude approximation for evaluating a practical dt in case of Re sufficiently high:

dt_p*(umax/dx+vmax/dy)<1 --> dt = alpha*dt_p (alpha = 0.1 - 0.5)

November 30, 2013, 10:18
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 Originally Posted by FMDenaro The numerical stability region (Reh,cfl) is determined by the type of numerical scheme you use, each one producing a different map of stability. I don't see your type of discretization, Re number and number of cells. I can suggest a rude approximation for evaluating a practical dt in case of Re sufficiently high: dt_p*(umax/dx+vmax/dy)<1 --> dt = alpha*dt_p (alpha = 0.1 - 0.5)
Yes, yes, my mistake. I am using CDS for all discretizations as per my instructions. My Reynolds number is only 100 (mu is 0.01, rho is 1, velocity is 1).

I'm not sure exactly what you mean by dt_p? And the value of alpha is dependent on Re?

 November 30, 2013, 10:26 #4 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,764 Rep Power: 71 what about the time-integration scheme?

 November 30, 2013, 10:40 #5 New Member   Join Date: Nov 2013 Posts: 5 Rep Power: 12 I am using an Explicit Time Advance scheme. I calculate the combination of the advective and viscous terms and its divergence from the initial velocity field. I solve Poisson's equation for pressure and then compute the velocity field at the next time step. I hope that answers your question.

November 30, 2013, 11:42
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Filippo Maria Denaro
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 Originally Posted by Atreides I am using an Explicit Time Advance scheme. I calculate the combination of the advective and viscous terms and its divergence from the initial velocity field. I solve Poisson's equation for pressure and then compute the velocity field at the next time step. I hope that answers your question.
If you are using a first order Explicit time integration, Re=100 will drive towards a diffusive stability constraint such as:

dt_d*(1/dx^2 + 1/dy^2)/Re <1/2.

As your sistem is non linear such value should be furhter reduced

 November 30, 2013, 11:52 #7 New Member   Join Date: Nov 2013 Posts: 5 Rep Power: 12 Well perhaps I have other errors in my code but this produces a far larger time step than I have been using and it is immediately unstable. From where did you derive this equation for stability? If I have a square that I have divided into a grid of 100 points with delta x = delta y = 0.1 that gives me a time step of 1 second per your equation?

November 30, 2013, 12:23
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Filippo Maria Denaro
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Quote:
 Originally Posted by Atreides Well perhaps I have other errors in my code but this produces a far larger time step than I have been using and it is immediately unstable. From where did you derive this equation for stability? If I have a square that I have divided into a grid of 100 points with delta x = delta y = 0.1 that gives me a time step of 1 second per your equation?

The constraint comes from the linear stabilty analysis of the parabolic equation dphi/dt=Gamma* Lap Phi, discretized with the FTCS scheme
Assuming a grid formed by 10x10 cells (dx=dy=0.1) you simply have
dt_d < 0.25. Therefore dt =0.1 should work.
If you see instability after few time-steps then there is an error in the code

November 30, 2013, 12:27
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 Originally Posted by FMDenaro The constraint comes from the linear stabilty analysis of the parabolic equation dphi/dt=Gamma* Lap Phi, discretized with the FTCS scheme Assuming a grid formed by 10x10 cells (dx=dy=0.1) you simply have dt_d < 0.25. Therefore dt =0.1 should work. If you see instability after few time-steps then there is an error in the code
I will investigate. Thanks very much for the help.

 Tags finite difference method, navier-stokes, stability