User:


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:(html file)

ID:(9156, 0)


Classical LBM solution in the BGK approach
Description

Variables
Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$f^{eq}_i$
feq_i
Componente $i$ de la distribución en equilibrio según BGK
-
$\rho$
rho
Densidad en el espacio
kg/m^3
$\vec{e}_i$
&e_i
Dirección de la componente $i$ según BGK
-
$c$
c
Factor de normalización de BGK
-
$\omega_i$
omega_i
Factor de peso en la componente $i$ según BGK
-
$f$
f
Función distribución de la teoría de transporte
-
$f^{(0)}$
f^0
Función distribución en equilibrio
-
$t$
t
Tiempo
s
$\tau$
tau
Tiempo de relajamiento
s
$\vec{u}$
&u
Velocidad en el espacio
m/s

Calculations

First, select the equation:   to ,  then, select the variable:   to 
Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

 Variable   Given   Calculate   Target :   Equation   To be used


Equations

Examples

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)

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 m is the mass of the particle, T the system temperature and k the Boltzmann constant.

(ID 8490)

En la aproximaci n Bhatnagar-Gross-Krook la distribuci n en equilibrio se asume como la de un gas de part culas sin interacci n

$f^{(0)}(\vec{x},\vec{v},t)=c(\vec{x},t)\left(\displaystyle\frac{m\beta}{2\pi}\right)^{3/2}e^{-\beta m(\vec{v}-\vec{u}(\vec{x},t))^2/2}$



con \vec{u} la velocidad del flujo, k la constante de Boltzmann, T la temperatura y m la masa de la particula. Si se desarrolla esta expresi n en el limite de velocidades \vec{u} comparada con la velocidad de las moleculas c\hat{e}_i se tiene que

$f_i^{eq}=\rho\omega_i\left(1+\displaystyle\frac{3\vec{u}\cdot\vec{e}_i}{c}+\displaystyle\frac{9(\vec{u}\cdot\vec{e}_i)^2}{2c^2}-\displaystyle\frac{3u^2}{2c^2}\right)$

con \omega_i los pesos dados por

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 distribuci n equilibrio cumpla las leyes de conservaci n.

(ID 9084)

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)

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)

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)

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)

In the case of particles of a liquid, the method LBM allows to develop simulators as shown in the example:(html file)

(ID 9156)

ID:(1153, 0)