I think that your result is quite logical. The Navier-Stokes equations for Poiseuille flow simplify to
\rho \nu \Delta u = \rho g
where \rho is density (as function of position), \nu is kinematic viscosity (constant throughout your channel), \Delta is the Laplacian operator, u is velocity, and g is gravitational acceleration or any other homogeneous acceleration (constant throughout your channel). You see that the density cancels and that the result MUST be the same as for a single phase fluid. Only if the kinematic viscosity differs, you would expect a modified velocity profile. You should choose different kinematic viscosities.
I am using a constant driving force for both phases (F) and I think applying the Chapman-Enskog expansion procedure to the lattice-Boltzmann equation for multiphase (SC Model), one obtains the following mass and momentum equations
for the fluid mixture treated as a single fluid
(\sum_a F_a )/(\rho) = - \nu \Delta u a=1, ... ,9 for 2D
where
\sum_a F_a = F_ext + F_{f-f} + F_{f-s}
where F_{ext}, F_{f-f} , and F_{f-s} are momenta contributed by external force, interaction between phases, and interaction of ĀÆfluid with solid, respectively. [Z.L. Yang et al. 2001]
In this case, density is not canceled out and fluids with different density should have different velocity profiles!
Furthermore, I see some peaks in velocity perpendicular to the direction of flow at the interface between fluid phases ( from Navier-Stokes, we know that it should be zero everywhere in the doamin ). I think this accounts for the existence of F_{f-f} at the interface. But it changes the velocity magnitude at points around that!
Do you know any solution for this problem?
I still think that the density cancels because F is momentum, and momentum is proportional to density. Please carefully recheck your equations and make sure that the difference between force and force density is clear.
Again: For two fluids with the same kinematic viscosity the velocity profile must always be a parabola in the presence of a homogeneous acceleration field. But acceleration is not force and also not force density!
I find it a bit strange that you have a velocity perpendicular to the main flow. Due to the symmetry, this would eventually lead to a compressibility of the fluid. Is your interface flat? It is a 2D simulation, right? Is your system symmetric with respect to the main velocity axis?
Hello, could you please tell me where did you get this formula?
nu_{Phys}=dt_{Phys}/(dx_{Phys}*dx_{Phys})*nu_{LB}
Even the dimensions do not match and thatās weird to me !
We have ālength^2ā vs. āsecondā (for dt) to be canceled and nu(Phys) dimension becomes equals to that of nu(LB).
Hello, i am preparing an algoritm for natural convection using the LBM. I have a square cavity of 1m * 1m. I have viscosity V_p and Prandtl number(Pr). where suffix āpā indicates in physical units and ālā indicates in lattice units. Now for LBM, i have set dx_l=dt_l=1 lattice unit, so C_l=1 lattice unit. Now i have 100 divisions in X and Y directions, so L_l=100 and also dx_p=dy_p=0.01m.
Now i have somewhere read that the non-dimensional number should be same for lattice units and physical units. In my case the Rayleigh number is 10^5. Now i have assemed the viscosity V_l=0.03 in lattice units. so, the value of alpha_l=Pr / V_l. Now i have found the value of g*beta_l=Ra * V_l * alpha_l / (delta_T * L_l^3).
Now, i want to see the temperature distribution and velocity distribution at a particular time, say 180 s. So, how many iteration i need to be performed?
Hello,
in order to get dt_p you need a quantity that contains time units. Here I guess itās you viscosity which has units [Length^2/Time]. You can deduce dt_p from the relation
V_p=V_l * dx_p^2/dt_p
Then the number of iteration is iter = 180 / dt_p.