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time step for incompressible flows

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  • 1 Post By Márcio Ricardo

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Old   August 22, 2005, 06:54
Default time step for incompressible flows
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I am working on incompressible flows and would like to know how time step based on convective and diffusive considerations are calculated. I consulted few papers, but there are not much details about this. I am not interested in the mere formula but also at how it was arrived.

Thanks in advance.

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Old   August 23, 2005, 00:01
Default Re: time step for incompressible flows
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I am trying to get the ans too. I am using factional step to solve the unsteady NS eqns. It's a semi implicit mtd ie Adam Bashford 2nd order for convective (explicit) & Crank nicolson 2nd order for diffusive (implicit). It was given in the paper that the CFL condition is

CFL=max(u(i)*delta_t/delta_x(i)) = 1.

So in order to solve the unsteady eqn, delta_t needs to be determined. However, u(i) is unknown. So how can we obtain delta_t?

Supposed we are simulating a lid driven cavity flow, we know that the max vel = 1. hence can we substitute into the CFL eqn to get delta_t?

Is that reasonable?
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Old   August 23, 2005, 08:34
Default Re: time step for incompressible flows
Márcio Ricardo
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At each time iteration you have a velocity field. u_i is the max absolute value that you get from the field you have just calculated. Another way to calculate your time step:

dt_conv = 1/(u/dx+v/dy) (1)

dt_difus = 1/(2*visc*(1/dx^2 + 1/dy^2)) (2)

dt = coef/(1/dt_conv + 1/dt_difus) (3)

where, in (1), u, v are the largest absolute velocity values you get in each time iteration.

The coef in (3) is a safety factor, usualy <=0.2

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