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Re: Force term – Units ?

#2369

Hi Robin,rnrnThank you for your response. However, I am still confuse about how it is implemented this body force.rnrnIf m** is the length in lattice units and t** is the time in lattice units. Then, according to the line of code for the poiseuille-forced-2d example:rn

Code:
T Ly = converter.numCells(ly);rn std::vector<T> poiseuilleForce(2,T());rn poiseuilleForce[0] = 8.*converter.getLatticeNu()*converter.getLatticeU() / (Ly*Ly);

rnThe poiseuilleForce x-component vector has acceleration lattice units (m**/(s**)²). In the poiseuille-forced-2d example the characteristics density and length are both unity. Thus, the “”mass”” in lattice units [CharRho * DeltaX³] is equal to 1. This is the reason why it is not wrote the “”mass”” in lattice units?rnrnThe body force F = rho*g [kg/m³ * m/s²] defined in Ghou et al. (2002) has not the units of N [kg * m/s²]. Thus, LBconverter.latticeForce(value) should not be used because the force in the physical system (value) will be dimensionless by LBconverter.physForce() which has N physical units. It is that right? rnrnIn the case of a incompressible and isothermal problem, the body force term g in the Navier-Stokes equations have [m/s²] physical units. This body force term in the dimensionless Navier-Stokes equations is:rn

Code:
g_d = 1/Fr² // Fr is the Froude number

rnTo pass from dimensionless to lattice units I use the same procedure presented in Latt (2008).rn

    g_d = dx/dt² * g_lbrng_lb = dt²/dx * g_drng_lb = dt²/dx * (1/Fr²)

rnI would like to know if it is correct to insert this gravity acceleration in lattice units [g_lb] in the force vector ? rnrnPD: Ghou et al. (2002) Discrete lattice effects on the forcing term in the lattice Boltzmann method.rnLatt (2008) Choice of units in lattice Boltzmann simulations.rnrnBest regards,rnrnAlejandro