on pressure in DFSPH

[twanverweij]

Dear mr. Bender and team,

I have some trouble recreating physical pressure in [Pa] from the DFSPH output and there is probably a normalization step I'm overlooking. Consider a unit box half filled with "water", i.e. rho=1000 [kg/m3]. The pressure at the bottom of the tank should be 5000 [Pa]. Solving with DFSPH however gives me the following P/rho^2. The bottom value is approximately 1e-4.

The P/rho^2 is therefore not P/((rho_i - rho_0)^2) (the compressibility divided by the density difference), but what does it represent instead, and how can I calculate back to SI units for pressure?

With gratitude in advance,
Twan

[twanverweij]

So.. after some further investigation into the normalization, I found that the P/rho^2 value is dependent on the time-step. Dividing the P/rho^2 number back by (dt^2) (reversing the normalization in the code) gives me the correct result when the unit of P/rho^2 is actually in the units of [Pa]/[kg/m^3] instead of [Pa]/([kg/m^3]^2) as the declaration seems to suggest. There's alot going on with density normalization (for multiphase?) so this could be the case, but I have to say I'm guessing here. I noticed too that for good results, the density error has to be small.

Any thoughts or ideas?