If your cylinder is stationary, then shouldn't the lift be zero theoretically? In that case, because your CL (or lift force) is small both experimentally and numerically, comparing the percentage difference between them is not very meaningful. Just my 5 cents. Cheers.

dear Minh Doan

thanks for your answer

my cylinder is stationary. and about lift & drag coefficient i have used root mean square and average values for comparing with experimental data

Theoretically, the flow around a circular cylinder will produce unsteady vortex street, both the lift coefficient and the drag coefficient oscillate with time. The time averaging method of transient calculation results will affect the final results, first of all, it is necessary to ensure the consistency between the numerical results processing method and the experimental processing method.

Secondly, the inflow conditions may have an impact, such as the turbulence intensity, dissipation ratio and other parameters should be consistent with the experimental results. And if the URANS method is used, it is better to use the convergence of the grid for numerical verification and validation, rather than grid independence.

If the RANS / LES hybrid method is used, it does not mean that the finer the mesh is, the better, the results of the calculation and the grid have no monotonous convergence results, that is, the mesh refinement can not bring more accurate results.

Other possible reasons include the grid distribution in the boundary layer, not just the grid refinement in the main flow region, whether the time step can satisfy the unsteady calculation and so on.

dear Yuan Tang

thanks for your perfect answer.

about experimental data which i want to reach. i use data of a article and i have not performed any experimental works.

furthermore i tried to make my simulation setup according to that article (some thing like dimensions, run time and turbulence intensity.)

about turbulence model i have used LES method.

about boundary layer mesh i think i have used a fine adequate mesh which can solve boundary layer correctly. because my drag coefficient is true.

dear Yuan Tang

do you know how might be the shape of drag coefficient- time graph at 3900 Reynolds ?

I'm very sorry, I haven't carried out any similar research, but you can refer to this paper"New insight on large-eddy simulation of flow past a circular cylinder at subcritical Reynolds number 3900(Article New insight on large-eddy simulation of flow past a circular...

dear Yuan Tang

Thank you for spending time with me!

i will check the paper you suggested certainly.