# Time step size for transient simulations of vortex shedding

I am simulating unsteady flow around a circular cylinder (using FLUENT). I am facing the following problem. Kindly please help me.

OBJECTIVE: To find out the Reynolds number at which vortex shedding actually starts during a flow around a circular cylinder.

Note: This study is already available in literature. Literature data show that vortex shedding starts at $\text{Re} = 46$ for a circular cylinder. I just want to redo that work.

Problem: I ran unsteady simulations from 45 to 50. But, I am not observing any vortex shedding. But, using the same setup, if I run simulations for $\text{Re} > 60$, I am observing vortex shedding and matching values of drag and St. Hence, I feel that I am making mistake in choosing the time step.

Method used: I came across several posts related to this and many methods to calculate the time step size.

1) From Strouhal number (0.2 for circular cylinder- approx)

$\text{Sr} (0.2) = (\text{frequency} \times \text{diameter}) / \text{velocity}$. From the above equation find the frequency. The total time period, $T = 1/\text{frequency}$. Hence, time step $\Delta t = T/25$ (approx).

2) From CFL condition $\Delta t = C_m X \Delta x / \text{velocity}$ where, $C_m = 1$ (approx) and $\Delta x$ = minimum cell size.

3) From domain length $\Delta t = \text{domain length} / (20 \times \text{velocity})$

I tried, all the above methods and I am not able to see vortex shedding between $\text{Re} = 45$ and $50$.

FLOW SCENARIO: Unsteady flow around a circular cylinder of $\text{dia} = 0.01\text{m}$ in the Reynolds number range of 45 to 50. The working fluid is water with $\text{viscosity} = 0.001003 \text{kg}/\text{ms}$ and $\text{density} = 998.2 \text{kg}/\text{m}^3$.

• What have you tried? The goal of "simulating unsteady flow" suggests a short time step is desired, but your post would greatly improved if you explained what time step you tried and in what way it did not give satisfactory results. – hardmath Feb 9 '16 at 11:56
• Hi, Thank you so much for trying to help me. I have modified my question. Please have a look at it. – Aarthy Feb 11 '16 at 3:51
• You have made the problem clearer and more interesting. Since you have raised the issue of a time step, I'm wondering what effects you will see if you reduce the time step, e.g. does separation occur at lower Reynolds numbers with shorter time steps? There seems little that I can do except shout encouragement. – hardmath Feb 11 '16 at 4:18
• I agree, this makes it clearer that this question is not really about FLUENT but about CFD. I have edited the title, tags and question a bit to make this clearer -- @Aarthy, if you don't agree, feel free to change back! – Christian Clason Feb 11 '16 at 8:21
• @Aarthy - If you think its due to the time step why don't you try a much smaller time step? What sort of discretization are you using, first order upwind, second order central; can numerical diffusion influence the critical value of Re where vortex shedding occurs? The literature study you mention, is it numerical or experimental? If experimental than many other factors may contribute to a lower critical Re than observed numerically. – nluigi Feb 21 '16 at 14:15

The physics you are trying to model exhibits parabolic and hyperbolic properties (some form of the Advection-Diffusion equation for calculating stream function or velocity) as well as elliptic properties in calculating vorticity. Let's assume a uniform grid ($\Delta x$=$\Delta y$=$h$). Because the physics is both hyperbolic and parabolic, the time step you should use is determined by the minimum time step calculated from the CFL for stability criteria for a hyperbolic equation:

$$c^{2} = \frac{(\sqrt{u^{2}+v^{2}})_\text{max}^{2}\Delta t^{2}}{h^{2}} \leq 2d$$

and from a parabolic equation in 2D:

$$d = \frac{\alpha \Delta t}{h^{2}} \leq \frac{1}{4}$$

where $\alpha$ is related to your Reynolds number, $c$ is your CFL value, and $\sqrt{u^{2}+v^{2}})_\text{max}^{2}$ is the maximum velocity over your domain, calculated at each time step.

It's important to note that these equations give you the maximum value you can use for the timestep. A smaller timestep won't hurt, especially if you're interested in the transient solution.

To address the problem you first explained, it's possible that your solution isn't set to a low enough error tolerance for your velocity or stream function calculation. The result of this would be more numerical dissipation and "smoothing" of your solution so that you wouldn't be able to see the smaller structures being formed at the lower Reynolds numbers.

I recommend to

1. Initialize your transient simulation with the solution of steady-state one as this will make the vortices appear faster
2. Activate the double precision in ANSYS fluent start dialog box

Whatever the literature available for flow past a cylinder, I think it is right. If you print the results in between your simulation, you will see the vortex shedding which is not exactly the same as with $\text{Re}=60$ but it is quite fluctuating and then along with increasing $\text{Re}$, its frequency increases and that is the reason to have vortex appearance at $\text{Re}>50$.

If you want to see the difference in vortex shedding, just output that into the file after every 5000 or 10000 iterations