Why use "Fourtakas (full)" in DDT?

Comments

• gfourtakas

  May 2020

  Hi Asalih3d,

  The density diffusion term (DDT) of Fourtakas et al., 2019 has been implemented in v5.0 of DualSPHysics. This version of the diffusion term has advantages over the traditional we have been using in DualSPHysics in the past (which is still present in the solver as DDT1).

  You have two options, either apply the diffusion term only to particles which do not interact with the fluid (that is 2h away from the boundary, DDT2) or apply the term in the whole (full DDT3) domain. Clearly the latter will produce much more accurate pressure field near the boundary free from numerical noise . Nevertheless, a severely truncated kernel with high density boundary particles from DBC will impact the accuracy of this diffusive term.

  For soothing lengths coefficients smaller than h/dp<1.5 the DDT works sufficiently well - even when only one line of boundary particles is used, for larger kernel support you may need to use more than one line of wall particles, this will help with the kernel truncation and most importantly the large density of the boundary particles.

  The kernels we use in DualSPHysics have a support of 2h thus, you can easily calculate how many particles are within the radius of your kernel and use the appropriate number of boundary particle lines.

  The message is simply a warning/reminder for the user.

  Happy DualSPHysics-v5-ing!!!

  Regards,

  George
• Asalih3d

  May 2020

  Thanks for the reply @gfourtakas !

  That makes sense, also the help file suggests heavily to use DDT3, so I will keep doing that. It would be nice though if the warning would explain (or just mention) that the criteria to look out for is; 2h <= x*Dp, where x is number of layers - since this is mentioned very nicely in the helpfile.

  Kind regards and thank you for the indepth explanation, made me understand it
• sph_tudelft_nl

  October 2020

  @gfourtakas Would you please double check my understanding?

  A) Re

  "1. The selected DDT '(Fourtakas et al 2019 (full)' needs several boundary layers when h/dp>1.5"

  am I right that boundary layer means layers of particles inside the boundary there? Just not to be led astray when speed reading.

  B) Re

  You have two options, either apply the diffusion term only to particles which do not interact with the fluid (that is 2h away from the boundary, DDT2) or apply the term in the whole (full DDT3) domain.

  As I understood the quote above, the DDT algorithm is exactly the same for the two options. The DSPH options decide where this method is applied.

  • With DDT3, the improved method is applied to all fluid particles.
  • With DDT2 the improved method is applied to all fluid particles except those within 2h of the solid wall. This region is a layer outside the boundary, like a canonical boundary layer.

  Is this reading correct?

  C) Does DDT2 mean that, in that layer outside the boundary, there is no falling back to any density diffusion term, not even the standard delta-SPH? So basically what's in place in that region? DDT0 (None) or DDT1 (Molteni)?

  D) Why do you mention large density of the boundary particles? In which way is it larger than that of the fluid particles? Do you mean density as rho, or density as number of particles inside the kernel? I presume the latter now, but it would be good to have your check.

  Thanks a lot for clarifying this. It helps me picture how to select the DDT and anticipate its effects.
• gfourtakas

  October 2020

  hi,

  On your questions:

  a) Yes, in SPH we (tend) to use particles to represent the physical boundary and complete the kernel support.

  b) correct.

  c) correct - DDT0 if DDT2 is used - I assume "... in that layer outside the boundary .." you mean the fluid particles near the BC and within the support domain of those BC particles.

  d) The fluid density (ρ).

  
  

  Thank you,

  George
• sph_tudelft_nl

  October 2020

  Thanks @gfourtakas. Proceeding pointwise as above:

  A) Clear. Let's recall that in flows the boundary layer is at the other side of the layers of solid particles in a boundary. Users still inexperienced in SPH may find the warning message confusing at first; my two cents.

  B) Thanks.

  C) Thanks. And we are on the same page.

  D) This escapes me still. Why the density of the boundary particles would be distinctively 'high'? If I recall it correctly, the density of boundary particles starts from RhopZero, the boundary-particle density is updated with the same mass-conservation equations as the fluid particles, and values larger than RhopZero are set to RhopZero. Is this keeping the boundary-particle density 'high' in the sense you were mentioning? Thanks in advance for integrating/differentiating.
• sph_tudelft_nl

  October 2020

  Please read "values larger than RhopZero are set to RhopZero" as "values smaller than RhopZero are set to RhopZero".