User:


Colisiones
Storyboard

>Model

ID:(1112, 0)


Colisiones
Description

Variables
Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$f_{in}$
f_in
Contribución a la función distribución que ingresan (gana)
-
$f_{out}$
f_out
Contribución a la función distribución que salen (pierde)
-
$f$
f
Función distribución de la teoría de transporte
-
$\sigma$
sigma
Sección eficaz de la colisión $(\vec{v}_1,\vec{v}_2\rightarrow\vec{v}'_1,\vec{v}'_2)$
m^2
$t$
t
Tiempo
s
$\tau$
tau
Tiempo de relajamiento
s
$v$
v
Velocidad de la partícula que afecta la distribución
m/s
$v_1$
v_1
Velocidad partícula 1 que colisiona
m/s
$v_21$
v_21
Velocidad partícula 1 que resulta de la colisión
m/s
$v_2$
v_2
Velocidad partícula 2 que colisiona
m/s
$v_22$
v_22
Velocidad partícula 2 que resulta de la colisión
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

In case the particles collide, the distribution function f(\vec{x},\vec{v},t) variert und\\n\\n

$\displaystyle\frac{df}{dt}\neq 0$



Collisions cause particles of neighboring cells to undergo a collision that takes them to the cell under consideration and particles within the cell being expelled. The first leads to an increase of f_{in} particles and the second to a f_{out} time loss \tau. Thus the Boltzmann transport equation with collisions can be written as

$\displaystyle\frac{df}{dt}=\displaystyle\frac{1}{\tau}(f_{in}-f_{out})$

(ID 9077)

In the case of collisions, two particles with velocity \vec{v}_1 and \vec{v}_2 collide to have velocities \vec{v}_1' and \vec{v}_2' respectively. The probability that the velocities after the collision are \vec{v}_1' and \vec{v}_2' can be estimated with the effective section \sigma that is\\n\\n

$\sigma(\vec{v}_1,\vec{v}_2\rightarrow\vec{v}_1',\vec{v}_2')d\vec{v}_1'd\vec{v}_2')$

\\n\\nAs the probability that the particles entering the collision are \vec {v}_1 and \vec{v}_2 are calculated with the distribution function\\n\\n

$f(\vec{x},\vec{v}_1,t)f(\vec{x},\vec{v}_2,t)$



and the displacement occurs as a function of the relative velocity |\vec{v}_2-\vec{v}_1|, it is finally found that the variation of the number of particles are

$f(\vec{x},\vec{v}_1,t)f(\vec{x},\vec{v}_2,t)|\vec{v}_2-\vec{v}_1|\sigma(\vec{v}_1,\vec{v}_2\rightarrow\vec{v}_12,\vec{v}_22)d\vec{v}_12d\vec{v}_22$

(ID 9078)

In the case of contributions to the cell, consider

$f(\vec{x},\vec{v}_1,t)f(\vec{x},\vec{v}_2,t)|\vec{v}_2-\vec{v}_1|\sigma(\vec{v}_1,\vec{v}_2\rightarrow\vec{v}_12,\vec{v}_22)d\vec{v}_12d\vec{v}_22$



Integrating on the speeds that initiate the collision and one of the resulting ones since the other is the contribution to the local distribution function

$\displaystyle\frac{1}{\tau}f_{in}(\vec{v})=\displaystyle\int d\vec{v}_1d\vec{v}_2d\vec{v}_12f(\vec{x},\vec{v}_1,t)f(\vec{x},\vec{v}_2,t)|\vec{v}_2-\vec{v}_1|\sigma(\vec{v}_1,\vec{v}_2\rightarrow\vec{v}_12,\vec{v})$

(ID 9079)

In the case that they leave the cell it is considered

$f(\vec{x},\vec{v}_1,t)f(\vec{x},\vec{v}_2,t)|\vec{v}_2-\vec{v}_1|\sigma(\vec{v}_1,\vec{v}_2\rightarrow\vec{v}_12,\vec{v}_22)d\vec{v}_12d\vec{v}_22$



Integrating on one of the speeds that initiate the collision and both resulting since the other is the contribution to the local distribution function

$\displaystyle\frac{1}{\tau}f_{out}(\vec{v})=\displaystyle\int d\vec{v}_1d\vec{v}_12d\vec{v}_22f(\vec{x},\vec{v}_1,t)f(\vec{x},\vec{v},t)|\vec{v}-\vec{v}_1|\sigma(\vec{v},\vec{v}_1\rightarrow\vec{v}_12,\vec{v}_22)$

(ID 9080)

With the term collisions that contribute

$f(\vec{x},\vec{v}_1,t)f(\vec{x},\vec{v}_2,t)|\vec{v}_2-\vec{v}_1|\sigma(\vec{v}_1,\vec{v}_2\rightarrow\vec{v}_12,\vec{v}_22)d\vec{v}_12d\vec{v}_22$



and those that reduce particles

$\displaystyle\frac{1}{\tau}f_{in}(\vec{v})=\displaystyle\int d\vec{v}_1d\vec{v}_2d\vec{v}_12f(\vec{x},\vec{v}_1,t)f(\vec{x},\vec{v}_2,t)|\vec{v}_2-\vec{v}_1|\sigma(\vec{v}_1,\vec{v}_2\rightarrow\vec{v}_12,\vec{v})$



you get the total exchange factor

$\displaystyle\frac{1}{\tau}(f_{in}-f_{out})=\displaystyle\int d\vec{v}_1d\vec{v}2d\vec{v}_12(f(\vec{x},\vec{v}2,t)f(\vec{x},\vec{v}_12,t)-f(\vec{x},\vec{v},t)f(\vec{x},\vec{v}_1,t))|\vec{v}-\vec{v}_1|\sigma(\vec{v},\vec{v}_1\rightarrow\vec{v}2,\vec{v}_12)$

(ID 9081)

ID:(1112, 0)