Which is the connection between residuals and under-relaxation factors (segregated solver)? I noticed that, if I set low under-relaxation factors, residuals drop down very quickly. Does the accuracy of the solution depend on the value of under-relaxation factors? Low values of such coefficients can mislead in judging convergence? Thanks a lot to everybody.

If you imagine a variable, say x-velocity then the new value of velocity as the computer is iterating is V(@n+1)=V(@n) + a*(delta V). Now you can see that if delta V is very big then this new value that the computer calculates, V(@n+1), is going to have a big change. If we make "a" have a value of say 0.5, then this large change in delta V will be halved and its influence halved. So then your residuals will reduce faster than if a=1, i.e. taking the full change of the x-velocity. BUT you will, if you leave it for long enough, reach the same answer. So if you are having convergence problems then a good tip is to reduce the under-relaxation of the momentum by 0.1 each time. You will notice a large increase in the ease your computer has at converging.

I know people who have been modelling highly viscous fluids with an under-relaxation for some variables at 0.01. It just takes much longer to converge, but does allow it to converge, i.e. the computer may not converge if the under-relaxation is at a higher value. You can also think of the under-relaxation as how hard you are driving the solver, values close to one are driving it hard and as you decrease the values the easier it is for you computer and the solver. Think of the residuals as an error, the lower the value the lower the error in your results.

Off course these explanations are rather simplified, but it should help you make a start with the program.

Hope this helps

Andy

Dear Andy

Thank you very much for reply. Your answer is very helpful.

Just a little question: If residuals are as an error, why do they decrease faster when the under-relaxation factors are low? If "a" is low, V(@n+1) should be almost the same as V(@n), shouldn't it? As a consequence, I think, the error on the solution should be almost the same. In some cases, setting under-relaxation to say 0.1 times the default value, residuals decrease of orders of magnitude in two or three iterations. Please, tell me where my misunderstanding is.

Thank you again for your kind help,

Ale

This is not a fob-of, but I would suggest you read the help menu in fluent, it is very good, and tells you what residuals are and for each solver, as I said before the description of error, is rather simplistic and it would be better if you get the full correct definition as fluent sees it.

Hope this helps

Andy

Thanks, and good luck to everybody.

Ale