Time step size and max iterations per time step

flotus1 · January 16, 2014, 07:18

I think so.
With larger time step, the solution evolves faster from the initial conditions towards the final solution.
But since you encountered difficulties with a steady-state solver, dont expect this to hold true for arbitrary time step sizes. The transient solver might have the same difficulties converging the solution when the time step size is too high.

Anna Tian · January 16, 2014, 07:24

Quote:

Originally Posted by flotus1

I think so.
With larger time step, the solution evolves faster from the initial conditions towards the final solution.
But since you encountered difficulties with a steady-state solver, dont expect this to hold true for arbitrary time step sizes. The transient solver might have the same difficulties converging the solution when the time step size is too high.

But will it be 10 times faster? I also encounter the problem of choosing large time step or small one now. Large time step need more iterations to converge at every time step. But it doesn't seem to evolve much faster from the initial conditions towards the final solution. It seems to me that, for this kind of 'stead' case, large time step can never be better than small one. Maybe it's better to have time step very small?

flotus1 · January 16, 2014, 07:35

If the flow is really time-dependent, you will get to the same point with 10 times less time steps if you choose a time step 10 times larger, provided both time step sizes are able to resolve the relevant transient effects.
Of course with an implicit method you will need more iterations per time step to achieve the same level of convergence.

But since you are using a transient solver as a computational workaround for a flow that might have no transient effects at all or at least we dont know the time scale of the flow, I would not expect the exact same behavior.

Lets have a look at a simple example, the couette flow between two parallel plates initially at rest.
The flow actually is time-dependent, but if the Reynolds number is sufficiently low, it will reach a steady state, a linear velocity profile between the two plates.
So we could simulate the long-term-solution of this flow both with a steady solver or with a transient solver.

But how long will it take for the transient solver? The viscous time scale at which the flow develops is
$T_{v} = \frac{L^2}{\nu}$
If we choose the time step size of the transient solver to be 1/10th of the viscous time scale, we need an order of 10-100 time steps to reach the steady state.
If we choose a time step size of 1/1000th of the viscous time scale, we need an order of 1000-10000 time steps to reach the same solution, which will obviously take longer.

To conclude: It is not always better to have a small time step size.

enass massoud · February 11, 2015, 07:58

I am working on 2 phase flow using vof method and i want to know how to calculate the time step and the number of time steps and how to determine the maximum number of iterations per time step?

ruben91 · April 13, 2015, 05:36

Hi everyone.

I am currently investigating multiphase flow in a pipe with regards to how it develops. I am currently using a grid size of 2 and the highest input velocity is 0.3m/s. In order to have my Courant number less than 1, I am using a fixed time step of 0.0006s.

However, I am getting a Global Courant number of 5.03. Would this affect the accuracy of my simulation? And if so how do I reduce the Global Courant number and is there even a need to do so?

Thank you.

villager · May 12, 2015, 21:04

Quote:

Originally Posted by ruben91

...
However, I am getting a Global Courant number of 5.03. Would this affect the accuracy of my simulation?

See this link
http://cape-forum.com/index.php/topi...2.html#msg1392
William provides guidelines for CFL numbers using VOF approach in FLUENT. They are also summarized here.

CFL number greater than some value (generally, 1, 2.5 for multi-stage FLUENT solver) would result in numerical instability when using explicit formulation. It is allowed for implicit solver, however, rising CFL leads to increased numerical error due to the fact that every mesh-based solution becomes less precious when increasing step size (both time step and mesh step, the latter just means coarsening the mesh). However, if you have no transient effects you can use higher CFL.
You can run several calculations refining the time-step, the results should converge to some values (it's like mesh convergence, just refining time-step instead of mesh element size).

moj167 · August 20, 2015, 13:39

I have two situations in unsteady state simulation of two phase flow in fluidized bed (gas-solid):
1- time step is 0.0001
2- time step is 0.001
in each case solution converge in each time step. in the case 1, after 10000 time step, solution is achieved for t=1 s. in the case 2, after 1000 time step, solution is achieved for t=1.
my question is:
two solution should be same for t=1?

flotus1 · August 20, 2015, 14:35

Only if the solution is independent of the time step size already for the larger time step.

moj167 · August 21, 2015, 04:38

Quote:

Originally Posted by flotus1

Only if the solution is independent of the time step size already for the larger time step.

thanks for your answer
how can in found that the solution is independent of time step?

flotus1 · August 21, 2015, 05:46

By doing almost exactly what you described

Run the same simulation with at least three different time step sizes. Compare the solutions, e.g. by plotting a variable of interest over the time step size.
The factor between the the different time step sizes does not have to be as large as 10. A factor of 2 should be sufficient.
"Pro" tip: if you want to know if the simulation is independent of the time step size for a time step size of lets say 0.01s. You don't need to run two additional simulations with even smaller time step sizes which can be a computationally expensive task.
You might as well run two additional simulations with larger time step sizes, e.g. 0.02s and 0.04s.

Talha490 · April 27, 2020, 11:59

with this much less time step you sure will get convergence easily and get better results but if you have to solve for like 80 seconds. It will take a lot of time. What you you do to solve in this situation?? And my second question is refining mesh and using low time step, does it help to get better results than using less fine mesh and time step of 1E-05

AidealZohary · October 23, 2020, 23:50

Hello Lucky Tran,

"As you decrease the time-step size, the residuals typically decrease faster. The initial guess to the next iteration uses the final solution from the previous time-step. Since the difference in physical time between time-steps is smaller, the difference in solution between the previous time-step and current time step is smaller and this causes the residuals to decrease faster when the time-step is shortened."

My case: I am simulating an airfoil at Re 1million using K omega sst intermittency. Please correct me if im wrong but...

From the highlighted statement,

My understanding is that if i want the lift and drag to converge faster (roughly), i should just maintain using a bigger time step because the change in solution is bigger. Later reduce the time step to increase accuracy (smaller changes in the solution).

I also saw your suggestion in another thread saying that 20 inner iteration is a good choice for each time step and it is better to run 20 iter x 2 timestep rather than 40 iter x 1 timestep.

I am somehow confused with this second statement, if the lift and drag can converge faster (rough values) using a bigger time step, should`nt we increase the total number of bigger timestep and once it settles down (somewhat constant) then use a smaller timestep.

Hope you can please clarify. Thank you!

p/s Just noticed theres a second page. Flotus1 I believe already answered this question

January 16, 2014, 07:35		#23
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	If the flow is really time-dependent, you will get to the same point with 10 times less time steps if you choose a time step 10 times larger, provided both time step sizes are able to resolve the relevant transient effects. Of course with an implicit method you will need more iterations per time step to achieve the same level of convergence. But since you are using a transient solver as a computational workaround for a flow that might have no transient effects at all or at least we dont know the time scale of the flow, I would not expect the exact same behavior. Lets have a look at a simple example, the couette flow between two parallel plates initially at rest. The flow actually is time-dependent, but if the Reynolds number is sufficiently low, it will reach a steady state, a linear velocity profile between the two plates. So we could simulate the long-term-solution of this flow both with a steady solver or with a transient solver. But how long will it take for the transient solver? The viscous time scale at which the flow develops is $T_{v} = \frac{L^2}{\nu}$ If we choose the time step size of the transient solver to be 1/10th of the viscous time scale, we need an order of 10-100 time steps to reach the steady state. If we choose a time step size of 1/1000th of the viscous time scale, we need an order of 1000-10000 time steps to reach the same solution, which will obviously take longer. To conclude: It is not always better to have a small time step size. Anna Tian likes this.

February 11, 2015, 07:58	time step size in vof method	#24
enass massoud Member enass Join Date: Feb 2015 Location: Alexandria-Egypt Posts: 30 Rep Power: 11	I am working on 2 phase flow using vof method and i want to know how to calculate the time step and the number of time steps and how to determine the maximum number of iterations per time step?

April 13, 2015, 05:36	Global Courant Number in Fluent 15.0	#25
ruben91 New Member Rubegan Join Date: Apr 2015 Posts: 1 Rep Power: 0	Hi everyone. I am currently investigating multiphase flow in a pipe with regards to how it develops. I am currently using a grid size of 2 and the highest input velocity is 0.3m/s. In order to have my Courant number less than 1, I am using a fixed time step of 0.0006s. However, I am getting a Global Courant number of 5.03. Would this affect the accuracy of my simulation? And if so how do I reduce the Global Courant number and is there even a need to do so? Thank you.

August 20, 2015, 13:39	Time step!	#27
moj167 New Member Mojtaba Rasteh Join Date: Aug 2015 Posts: 2 Rep Power: 0	I have two situations in unsteady state simulation of two phase flow in fluidized bed (gas-solid): 1- time step is 0.0001 2- time step is 0.001 in each case solution converge in each time step. in the case 1, after 10000 time step, solution is achieved for t=1 s. in the case 2, after 1000 time step, solution is achieved for t=1. my question is: two solution should be same for t=1?

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January 16, 2014, 07:18		#21
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	I think so. With larger time step, the solution evolves faster from the initial conditions towards the final solution. But since you encountered difficulties with a steady-state solver, dont expect this to hold true for arbitrary time step sizes. The transient solver might have the same difficulties converging the solution when the time step size is too high.

August 20, 2015, 14:35		#28
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	Only if the solution is independent of the time step size already for the larger time step.

August 21, 2015, 05:46		#30
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	By doing almost exactly what you described Run the same simulation with at least three different time step sizes. Compare the solutions, e.g. by plotting a variable of interest over the time step size. The factor between the the different time step sizes does not have to be as large as 10. A factor of 2 should be sufficient. "Pro" tip: if you want to know if the simulation is independent of the time step size for a time step size of lets say 0.01s. You don't need to run two additional simulations with even smaller time step sizes which can be a computationally expensive task. You might as well run two additional simulations with larger time step sizes, e.g. 0.02s and 0.04s.

April 27, 2020, 11:59		#31
Talha490 New Member Talha Bin Tahir Join Date: Mar 2020 Posts: 1 Rep Power: 0	with this much less time step you sure will get convergence easily and get better results but if you have to solve for like 80 seconds. It will take a lot of time. What you you do to solve in this situation?? And my second question is refining mesh and using low time step, does it help to get better results than using less fine mesh and time step of 1E-05

October 23, 2020, 23:50		#32
AidealZohary New Member Aideal Zohary Join Date: Feb 2019 Location: Malaysia Posts: 28 Rep Power: 7	Hello Lucky Tran, "As you decrease the time-step size, the residuals typically decrease faster. The initial guess to the next iteration uses the final solution from the previous time-step. Since the difference in physical time between time-steps is smaller, the difference in solution between the previous time-step and current time step is smaller and this causes the residuals to decrease faster when the time-step is shortened." My case: I am simulating an airfoil at Re 1million using K omega sst intermittency. Please correct me if im wrong but... From the highlighted statement, My understanding is that if i want the lift and drag to converge faster (roughly), i should just maintain using a bigger time step because the change in solution is bigger. Later reduce the time step to increase accuracy (smaller changes in the solution). I also saw your suggestion in another thread saying that 20 inner iteration is a good choice for each time step and it is better to run 20 iter x 2 timestep rather than 40 iter x 1 timestep. I am somehow confused with this second statement, if the lift and drag can converge faster (rough values) using a bigger time step, should`nt we increase the total number of bigger timestep and once it settles down (somewhat constant) then use a smaller timestep. Hope you can please clarify. Thank you! p/s Just noticed theres a second page. Flotus1 I believe already answered this question