1
$\begingroup$

I am studying fluid-structure interaction across integrated circuits packaging by using Ansys Fluent (I'm a beginner). Currently I had an issue about the time step size.

When I set time step size of 1 s it takes about more than 100 s for a complete fluid flow. While for time steps of 0.1 s and 0.01 s takes about 50 s and 10 s for a complete flow.

My question is, how does time step size affect fluid flow?

Let say for the 10 s case, $$(\text{time step size})\times(\text{no. of time step})$$ $$1.0 \text{ s} \times 10 = 10 \text{ s}$$ $$0.1 \text{ s} \times 100 = 10 \text{ s}$$ $$0.01 \text{ s} \times 1000 = 10 \text{ s}$$

I suppose, at specific time (let say at $t=10\text{ s}$) for whatever time step size I'm using, I'll be getting same (or similar) position of flow. Am I wrong?

Case details: - VOF multiphase - Implicit solver - Transient

Thanks in advance.

Improve this question
$\endgroup$
1
  • $\begingroup$ How did I select the time step size of 0.0002 s as the datum? Have I can conduct a time independence study as well? $\endgroup$
    – Guest
    Commented Jun 25, 2020 at 21:37

2 Answers 2

Reset to default
1
$\begingroup$

You've given scant details for your problem, so I can only quote general trends.

Time step size affects numerical error and stability. If you pick a really large step size, your results may be meaningless because the resulting numerical method is unstable. You'll typically see large oscillations in the fluid flow field, and cutting the time step will be necessary to obtain a stable numerical method.

Assuming that your time step is smaller than the time step governed by the stability limit for your problem and choice of ODE solver, cutting the time step further will generally decrease the numerical error in your solution, so there is a tradeoff between the magnitude of the error you are willing to accept in your calculated solution and the computational effort you will need to expend to calculate a solution such that the error is of that acceptable magnitude (or less).

So, at a specific time for whatever time step size you're using, you may or may not get similar flow fields, depending on stability and accuracy of the numerical method for your particular choice of step size and your problem. A good thing to do would be to run a convergence study of your problem for multiple choices of time step and compare the results, using the calculated solution at the smallest choice of time step as the "true solution". If you see large differences as you approach your smallest time step solution, something is wrong, and you will need to diagnose the error; cutting the time step might help, but there could be other problems.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Implicit solvers are unconditionally stable, so regardless of OP's time step the simulation will be stable. However, accuracy will suffer with increased time steps. $\endgroup$
    – user7257
    Commented Nov 3, 2014 at 16:56
  • $\begingroup$ @user7251 Not all implicit solvers are unconditionally stable. For example, you can show that there is no A-stable BDF higher than 6th order. $\endgroup$
    – Aurelius
    Commented Nov 3, 2014 at 17:02
  • $\begingroup$ @Aurelius fair enough, I should have said that they are typically unconditionally stable. $\endgroup$
    – user7257
    Commented Nov 3, 2014 at 17:17
  • 1
    $\begingroup$ The second Dahlquist barrier is more severe: there are no A-stable linear multistep methods with order greater than 2. I didn't want to make any assumptions about the order of the discretization; if pressed, yes, usually for engineering-type simulations, low-order spatial and temporal discretizations are used. In that case, stability probably won't be an issue; I talk about it anyway for pedagogical reasons. $\endgroup$
    – Geoff Oxberry
    Commented Nov 3, 2014 at 19:18
  • 1
    $\begingroup$ @Aurelius Your statement "there is no A-stable BDF higher than 6th order" is true since there is no A-stable BDF of order higher than 2. But, what you actually meant is probably that the BDF methods up to 6th order are stiff-stable (sometimes called A(alpha)-stable). $\endgroup$
    – Tobias
    Commented Dec 4, 2014 at 7:29
0
$\begingroup$

What do you mean by "complete flow"? If you're saying that after that amount of physical time, the flow solution in your transient solver becomes steady state then it's because you're not actually resolving the unsteady transients in the flow with your timestep selection. If it has a steady state, altering the time-step of an implicit method like you are is actually more akin to a relaxation parameter on a Newton method than it is actual time-stepping.

If you're only interested in the steady-state solution, then don't use the transient solver, use the steady one.

If you are actually interested in the transient / start-up solution, then do a time-step convergence study with continually decreasing time steps until you reach a point where the flow is the same for any smaller timestep.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.