Hi,

I've been testing my NS code for different Re of cavity flow.

My range is from 100 to 10000 (as Ghia's)

Since mine is a transient solver, I'm not sure when I should stop.

It seems that as Re increase, the non-dimensional time to reach steady state also increases. Is that so?

Is there any recommended time for different values of Re?

Also, my u,v results compared to Ghia's is about 5% difference. Is that acceptable? I'm using a 2nd order projection mtd.

Thanks alot!

For each Re, at different time u will be the converged solution. so You can check for the rms error in the previous and current time step velocities (say rmsU, rmsV, rmsW) and then find out the maximum error: rms_max(rmsU,rmsV,rmsV) check if (rms_max/dt) > tolerance I kept the tolerance as 1e-6 and got the approximately same values as Ghia et al. I think 5% is acceptable.

I am unable to accept that 5%, because when I talked to some gurus I was told that <1% deviation may be accepted.It is not the time, it is acceptable error level which determines the convergence.non-dim residue of continuity eqn (Patankar's b) <1e-04 is recommended by Spalding.with increased Re, at the top of the driven cavity the diffusion becomes too low and as a result the convection is to be arrested. a refined grid helps. but if you follow uniform gridding, better to consider some different discretisation for the convection term.

To really answer your question, I'd suggest you experiment with different times, different mesh densities, etc. Keep track of the convergence parameters, tabulate them, and try to draw your own conclusions.

The driven cavity is not a 'real' flow, but it's a very good problem to build your intuition or 'feel' for how your code works.

You need to do this problem enough different ways that you can decide for yourself what's 'good enough.'

hi ramp,

just to confirm, is the rms error formula = sqrt( sum(u(new)^2-u(old)^2)/n ) ?

i'll then get 1 rms error value each for u,v,w. Then the largest of these 3 values will be the rms_max and rms_max/dt must be < 1e-6 to stop the run.

Is that correct?

Thanks!

Yes Quarkz, Its correct.

hi,

just to confirm again,

is it sqrt( sum(u(new)^2-u(old)^2)/n ) or sqrt( sum((u(new)-u(old))^2)/n )

thanks again

Hi Quarkz, Its second one... I am making it more clear

rms_u=(U_new-U_old)^2

rms_v=(V_new-W_old)^2

rms_w=(W_new-W_old)^2

N=Ni*Nj*Nk

err_u=sqrt(sum of rms_u/N)

err_v=sqrt(sum of rms_v/N)

err_w=sqrt(sum of rms_w/N)

max_rms=maximum of (err_u,err_v,err_w)/dt

------ Check for convergence -------

If max_rms < 1.0e-6 ------> STOP

Have a nice time..... ramp

Hi,

thank you so much!