# Setting Time step for Implicit unsteady

 Register Blogs Members List Search Today's Posts Mark Forums Read

 April 26, 2020, 16:18 Setting Time step for Implicit unsteady #1 New Member   Join Date: Apr 2020 Posts: 10 Rep Power: 2 Hi all, I want to simulate flow over a bump using STARCCM+ for unsteady flow. In paper its given that time step is 10000. Since I am new to starccm. I don't know how to set it. - In solver, under implicit unsteady it given that that default time step 0.001s what does it mean ? -Stopping criteria: Maximum physical time given has 5s and Maximum inner iteration has 5. I want convergence continuity, momentum, energy and SA as below 1e-5. How to incorporate 10000 time step or more to get convergence below 1e-5. If I run for default functions convergence does not go below 1e-3.

 April 27, 2020, 02:19 #2 Senior Member   Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 4,017 Rep Power: 48 If it's not clear what a time-step size of 0.001 s is in an unsteady simulation, then you need serious help. By default there are no stopping criteria for residuals. And there shouldn't be, because they are not a measure of convergence. But if you still want them, you have to create stopping criteria for residuals. Right click on stopping criteria and create them from the residual monitors. You will probably need to increase the maximum number of inner iterations to a larger number (999999 will work) to ensure the residual criteria is reached. Set the maximum physical time based on the time-step size and number of time-steps you want to run.

April 27, 2020, 02:35
#3
New Member

Join Date: Apr 2020
Posts: 10
Rep Power: 2
Quote:
 Originally Posted by LuckyTran If it's not clear what a time-step size of 0.001 s is in an unsteady simulation, then you need serious help. By default there are no stopping criteria for residuals. And there shouldn't be, because they are not a measure of convergence. But if you still want them, you have to create stopping criteria for residuals. Right click on stopping criteria and create them from the residual monitors. You will probably need to increase the maximum number of inner iterations to a larger number (999999 will work) to ensure the residual criteria is reached. Set the maximum physical time based on the time-step size and number of time-steps you want to run.
time step is delta"t" which we discritize based on time marching in Control volume because of unsteadiness. My question is why it is default set 0.001s. If I keep time step as 1s and physical maximum physical time to 10000. It is 10000 time step. So what's the difference between keeping 0.001s and 1s?

What is the purpose of inner iteration? If I keep inner iteration 1 and simulate for 10000 time step and equations convergers to required convergence criteria (ex:1e-5).

Is that during each step inner iteration should converge to required convergence criteria?

April 27, 2020, 09:46
#4
Senior Member

Lucky Tran
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 4,017
Rep Power: 48
Quote:
 Originally Posted by ranjithdj What is the purpose of inner iteration? If I keep inner iteration 1 and simulate for 10000 time step and equations convergers to required convergence criteria (ex:1e-5).
Who even does this? You don't check your residuals only at the last time-step.... You must check them every time-step. Are you confusing unsteady simulations with a steady simulation?

The difference between 0.001s and 1.0 s is 0.999 seconds. I don't know what to say beyond that. The default number has to be something and in Star it is 0.001s. You can use whatever deltaT is appropriate for your problem. If you know what temporal discretization is then you should have an idea of the implications of running a case bigger/smaller deltaT.

We use iterative solvers here, for a variety of reasons. For each time-step of 0.001s, you have to solve many iterations before your calculation is converged for that time-step. If you set the inner iterations to 1, you will not only not converge, it is also unlikely you will ever reach your convergence criteria. Yes you need to converge every time-step.

April 27, 2020, 14:59
#5
New Member

Join Date: Apr 2020
Posts: 10
Rep Power: 2
Quote:
 Originally Posted by LuckyTran Who even does this? You don't check your residuals only at the last time-step.... You must check them every time-step. Are you confusing unsteady simulations with a steady simulation? The difference between 0.001s and 1.0 s is 0.999 seconds. I don't know what to say beyond that. The default number has to be something and in Star it is 0.001s. You can use whatever deltaT is appropriate for your problem. If you know what temporal discretization is then you should have an idea of the implications of running a case bigger/smaller deltaT. We use iterative solvers here, for a variety of reasons. For each time-step of 0.001s, you have to solve many iterations before your calculation is converged for that time-step. If you set the inner iterations to 1, you will not only not converge, it is also unlikely you will ever reach your convergence criteria. Yes you need to converge every time-step.
Yes I have confusion between steady and unsteady. I used to work with steady condition. This time I am working in unsteady state. In CFD class I discretized the unsteady condition. But in practical, I didn't use it. Im referring some journal and they mentioned only 10000 time step. So I don't know how to take this time step

June 13, 2020, 02:43
#6
New Member

Join Date: Dec 2017
Posts: 7
Rep Power: 4
Quote:
 Originally Posted by ranjithdj Yes I have confusion between steady and unsteady. I used to work with steady condition. This time I am working in unsteady state. In CFD class I discretized the unsteady condition. But in practical, I didn't use it. Im referring some journal and they mentioned only 10000 time step. So I don't know how to take this time step
The journal article might have meant simulation running for a total of 10000 time steps. Time step size is different. Suppose you select the time step size of 0.0001s, then you are performing calculations for (0.0001 x 10,000 = ) 1.0 s.