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Relaxation Approach
Equation
One way to solve Boltzmann's general equation is to linearize the equation by assuming that the collision term can be written as the difference between the distribution function and the equilibrium solution represented by the distribution function of Maxwell Boltzmann
| $\displaystyle\frac{df}{dt}=-\displaystyle\frac{1}{\tau}(f-f^{(0)})$ |
ID:(9083, 0)
Distribution in Balance (Gas of Particles)
Equation
The equilibrium distribution can be approximated by a distribution of Maxwell Boltzmann
| $f_i^{eq}=\displaystyle\frac{m}{2\pi kT}e^{-m|c\vec{e}_i-\vec{u}|^2/2kT}$ |
Where
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Bhatnagar-Gross-Krook Approach
Equation
En la aproximación Bhatnagar-Gross-Krook la distribución en equilibrio se asume como la de un gas de partículas sin interacción
| $f^{(0)}(\vec{x},\vec{v},t)=c(\vec{x},t)\left(\displaystyle\frac{m\beta}{2\pi}\right)^{3/2}e^{-\beta m(\vec{v}-\vec{u}(\vec{x},t))^2/2}$ |
con
| $f_i^{eq}=\rho\omega_i\left(1+\displaystyle\frac{3\vec{u}\cdot\vec{e}_i}{c}+\displaystyle\frac{9(\vec{u}\cdot\vec{e}_i)^2}{2c^2}-\displaystyle\frac{3u^2}{2c^2}\right)$ |
con
| Modelo | $\omega_i$ | Index |
| 1DQ3 | ? | i=0 |
| - | ? | i=1, 2 |
| 2DQ9 | 4/9 | i=0 |
| - | 1/9 | i=1,...,4 |
| - | 1/36 | i=5,...,8 |
| 3DQ15 | 1/3 | i=0 |
| - | 1/18 | i=1,...,6 |
| - | 1/36 | i=7,...,14 |
| 3DQ19 | ? | i=0 |
| - | ? | i=1,...,6 |
| - | ? | i=7,...,18 |
que se determinan asegurando que la distribución equilibrio cumpla las leyes de conservación.
ID:(9084, 0)
Example of Density
Description
In D2Q9 the density is calculated simply by adding the different factors
| $\rho(\vec{x},t) = m\displaystyle\int f(\vec{x},\vec{v},t)d\vec{v}$ |
\\n\\nso you have\\n\\n
$rho[x,y] = O[x,y]+N[x,y]+E[x,y]+S[x,y]+W[x,y]+NE[x,y]+SE[x,y]+NW[x,y]+SW[x,y]$
ID:(9152, 0)
Example of Speed in x
Description
In D2Q9 the flow rate is calculated simply by adding the different factors
| $\vec{u}(\vec{x},t) = \displaystyle\frac{m}{\rho}\int \vec{v}f(\vec{x},\vec{v},t)d\vec{v}$ |
so you have
```
u_x[x,y] = E[x,y] + NE[x,y] + SE[x,y] - W[x,y] - NW[x,y] - SW[x,y]
```
ID:(9153, 0)
Example of Speed in y
Description
In D2Q9 the flow rate is calculated simply by adding the different factors
| $\vec{u}(\vec{x},t) = \displaystyle\frac{m}{\rho}\int \vec{v}f(\vec{x},\vec{v},t)d\vec{v}$ |
so you have
```
u_y[x,y] = N[x,y] + NE[x,y] + NW[x,y] - S[x,y] - SE[x,y] - SW[x,y]
```
ID:(9154, 0)
Ejemplo de elemento de Colisión
Description
In case D2Q9 the term collision is calculated by summing the different factors
| $f_i^{eq}=\rho\omega_i\left(1+\displaystyle\frac{3\vec{u}\cdot\vec{e}_i}{c}+\displaystyle\frac{9(\vec{u}\cdot\vec{e}_i)^2}{2c^2}-\displaystyle\frac{3u^2}{2c^2}\right)$ |
so you have for each cell
```
O = O+w(4rho/9)(1-3u2/2) - O)
E = E+w(rho/9)(1 + u_x/3+5u_x^2-3u2/2)-E)
W = W+w(rho/9)(1 - u_x/3+5u_x^2-3u2/2)-W)
N = N+w(rho/9)(1 + u_y/3+5u_y^2-3u2/2)-N)
S = S+w(rho/9)(1 - u_y/3+5u_y^2-3u2/2)-S)
NE = NE+w(rho/36)(1+u_x/3+u_y/3+5(u2+2u_xu_y)/2-3u2/2) - NE)
SE = SE+w(rho/36)(1+u_x/3-u_y/3+5(u2-2u_xu_y)/2-3u2/2) - SE)
NW = NW+w(rho/36)(1-u_x/3+u_y/3+5(u2-2u_xu_y)/2-3u2/2) - NW)
SW = SW+w(rho/36)(1-u_x/3-u_y/3+5(u2+2u_xu_y)/2-3u2/2) - SW)
```
with
```
u2 = u_x^2+u_y^2
```
ID:(9155, 0)
Example Hydrodynamic Simulator
Description
In the case of particles of a liquid, the method LBM allows to develop simulators as shown in the example:
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