Boltzmann Equation

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Another approach is to look at the molecular level of fluids from a statistical standpoint (Rice 2006b). From this standpoint the independent variables are space, velocity and time, while the dependent variable is a molecular distribution function for species $i$ , ${{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right)$ . The Boltzmann equation relates the distribution function at $\left( \mathbf{x},\mathbf{c},t \right)$ to the distribution function at $(\mathbf{x}+\Delta \mathbf{x},$ $\mathbf{c}+\Delta \mathbf{c},$ $t + Δ t).$

The location in space is x, and the particle velocity is c. It is important to note that the particle velocity is directly related to the mass average velocity, V, that is used throughout this book. This distribution function can be related to the Navier-Stokes equations as well as other transport equations; these relationships give insight to the origin of transport coefficients such as viscosity. The species distribution function is related to the number of particles, $N i$ , at a location less than $x 0$ , and with a velocity less than $c 0$ at time $t$ .

$\begin{align} & {{N}_{i}}=particles\left( \mathbf{x}<{{\mathbf{x}}_{0}},\mathbf{c}<{{\mathbf{c}}_{0}},t \right) \\ & {{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right)=\frac{{{\partial }^{6}}{{N}_{i}}}{\partial {{\mathbf{x}}_{1}}\partial {{\mathbf{x}}_{2}}\partial {{\mathbf{x}}_{3}}\partial {{\mathbf{c}}_{1}}\partial {{\mathbf{c}}_{2}}\partial {{\mathbf{c}}_{3}}}=\frac{{{\partial }^{6}}{{N}_{i}}}{{{\partial }^{3}}\mathbf{x}{{\partial }^{3}}\mathbf{c}} \\ \end{align}\qquad \qquad( 1 )$

For simplicity, the partial derivatives with respect to all spatial directions and all velocity components are written as ${{\partial }^{3}}\mathbf{x}=\partial {{\mathbf{x}}_{1}}\partial {{\mathbf{x}}_{2}}\partial {{\mathbf{x}}_{3}}$ and ${{\partial }^{3}}\mathbf{c}=\partial {{\mathbf{c}}_{1}}\partial {{\mathbf{c}}_{2}}\partial {{\mathbf{c}}_{3}}$ , respectively. The number density of particles, $n i$ , is the average number of particles of species $i$ per unit volume.

${{n}_{i}}\left( \mathbf{x},t \right)=\int{{{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right){{d}^{3}}\mathbf{c}}\qquad \qquad( 2 )$

The limits of the integral are not shown because it incorporates all the velocities from $-\infty$ to $\infty$ . From the distribution function, macroscopic flow variables such as density $\left( \rho \right)$ , the mean species velocity ( $V i$ ), the mass mean velocity (V), the average kinetic energy $(\bar{e})$ and the average random kinetic energy or internal energy (e) can be defined.

${{\rho }_{i}}=\int{{{m}_{i}}{{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right){{d}^{3}}\mathbf{c}}\qquad \qquad( 3 )$

${{\rho }_{i}}{{\mathbf{V}}_{i}}=\int{{{m}_{i}}\mathbf{c}{{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right){{d}^{3}}\mathbf{c}}\qquad \qquad( 4 )$

$\rho \mathbf{V}=\sum\limits_{i}{\int{{{m}_{i}}\mathbf{c}{{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right){{d}^{3}}\mathbf{c}}}\qquad \qquad( 5 )$

$\rho \bar{e}=\sum\limits_{i}{\frac{1}{2}\int{{{m}_{i}}{{\mathbf{c}}^{2}}{{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right)}{{d}^{3}}\mathbf{c}}\qquad \qquad( 6 )$

$\rho e=\sum\limits_{i}{\frac{1}{2}\int{{{m}_{i}}\mathbf{w}_{i}^{2}{{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right)}{{d}^{3}}\mathbf{c}}\qquad \qquad( 7 )$

The mass of a single particle is $m i$ . The random velocity $\left( {{\mathbf{w}}_{i}} \right)$ for species $i$ is defined as the difference between a molecule’s actual velocity and the mass mean velocity.

${{\mathbf{w}}_{i}}=\mathbf{c}-\mathbf{V}\qquad \qquad( 8 )$

The temperature is a function of the average random kinetic energy.

$\frac{3}{2}{{k}_{B}}T=\frac{\rho }{n}e\qquad \qquad( 9 )$

It is important to note that the average kinetic energy and average random kinetic energy neglect both rotational and vibrational kinetic energy and are only accurate in cases where these energies are small compared to translational energy, as is the case for a monatomic ideal gas. Since this description is the most accurate for a monatomic gas, the terms molecule and particle are interchangeable. Also, for the following discussion, all the gas particles are assumed to have a constant mass.

As is discussed above, the Boltzmann equation relates the distribution function at (x, c, t) to the distribution function at a projected position ( $x + Δ x, c + Δ c, t + Δ t$ ). The change in spatial position and velocity are related to the particle velocity and particle acceleration (a), which is the external body force exerted on a particle such as gravity or a Lorentz force.

$\Delta \mathbf{x}=\mathbf{c}\Delta t\qquad \qquad( 10 )$

$\Delta \mathbf{c}=\mathbf{a}\Delta t\qquad \qquad( 11 )$

The Boltzmann equation for species $i$ is (Chapman and Cowling, 1970):

$\left( {{f}_{i}}\left( \mathbf{x}+\mathbf{c}\Delta t,\mathbf{c}+\mathbf{a}\Delta t,t+\Delta t \right)-{{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right) \right){{d}^{3}}\mathbf{x}{{d}^{3}}\mathbf{c}={{\Omega }_{i}}\left( {{f}_{i}} \right){{d}^{3}}\mathbf{x}{{d}^{3}}\mathbf{c}dt\qquad \qquad( 12 )$

The collision term, the rate of change of $f i$ due to collisions, is denoted by $Ω i (f i)$ . This function is discussed in more detail below. Expanding the first term by Taylor series yields:

${{f}_{i}}\left( \mathbf{x}+\mathbf{c}\Delta t,\mathbf{c}+\mathbf{a}\Delta t,t+\Delta t \right)={{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right)+\left( \frac{\partial {{f}_{i}}}{\partial t}+\nabla {{f}_{i}}\cdot \mathbf{c}+{{\nabla }_{\mathbf{c}}}{{f}_{i}}\cdot \mathbf{a} \right)\Delta t\qquad \qquad( 13 )$

The gradient operators without a subscript and with a subscript c are $\nabla =\left[ \begin{matrix} \frac{\partial }{\partial {{\mathbf{x}}_{1}}} & \frac{\partial }{\partial {{\mathbf{x}}_{2}}} & \frac{\partial }{\partial {{\mathbf{x}}_{3}}} \\ \end{matrix} \right]$ and ${{\nabla }_{\mathbf{c}}}=\left[ \begin{matrix} \frac{\partial }{\partial {{\mathbf{c}}_{1}}} & \frac{\partial }{\partial {{\mathbf{c}}_{2}}} & \frac{\partial }{\partial {{\mathbf{c}}_{3}}} \\ \end{matrix} \right]$ , respectively. Therefore, the Boltzmann equation can be rewritten as:

$\frac{D{{f}_{i}}}{Dt}=\frac{\partial {{f}_{i}}}{\partial t}+\mathbf{c}\cdot \nabla {{f}_{i}}+\mathbf{a}\cdot {{\nabla }_{\mathbf{c}}}{{f}_{i}}={{\Omega }_{i}}\left( f \right)\qquad \qquad( 14 )$

If no collisions occurs, the Boltzmann equation states that the distribution function does not change as it moves along the trajectory of a particle, $D f i / D t = 0$ . In most real systems, however, collisions do occur. Particles exhibit long-range and short-range forces. When the long-range forces are negligible, as is the case for a neutral gas, the particles are assumed to interact only when the short-range forces are prevalent. The short-range forces are prevalent when the distance between the particles is on the order of the particle diameter. Since the particles are this close to one another only for a very short time, the interaction is treated as a collision. Also, it is assumed that only binary collisions exist, therefore the higher order collisions, ${{\Omega }_{i}}({{f}_{i}}{{f}_{i}}),\text{ }\Omega ({{f}_{i}}{{f}_{i}}{{f}_{i}}),\text{ }\Omega ({{f}_{i}}{{f}_{i}}{{f}_{i}}{{f}_{i}})\ldots$ are not considered. This assumption is not good for dense gases or fluids. The collision between two particles with velocities $\mathbf{c}$ and ${{\mathbf{c}}_{A}}$ before collision results in the particles having velocities $\mathbf{{c}'}$ and ${{\mathbf{{c}'}}_{A}}$ after collision. The particles are assumed to retain their mass before and after collision, therefore the conservation of momentum and energy require:

$m\mathbf{c}+{{m}_{A}}{{\mathbf{c}}_{A}}=m\mathbf{{c}'}+{{m}_{A}}{{\mathbf{{c}'}}_{A}}\qquad \qquad( 15 )$

$\frac{1}{2}m{{\mathbf{c}}^{2}}+\frac{1}{2}{{m}_{A}}\mathbf{c}_{A}^{2}=\frac{1}{2}m{{\mathbf{{c}'}}^{2}}+\frac{1}{2}{{m}_{A}}\mathbf{{c}'}_{A}^{2}\qquad \qquad( 16 )$

Also, the collisions are considered elastic, therefore:

$\left| \mathbf{c}-{{\mathbf{c}}_{A}} \right|=\left| \mathbf{{c}'}-{{{\mathbf{{c}'}}}_{A}} \right|\qquad \qquad( 17 )$

Figure 1: Elastic Collision of Two Particles

Squaring eqs, (14) and (16) and adding them together yields the energy equation, (15). The impact of two particles is presented in Fig. 1. Since the collisions are frictionless, momentum and energy are only transferred in the normal direction (along the line of impact). Also, the impact occurs on a single plane; therefore, the relative velocity of the two particles can be deflected by an angle $θ$ while the magnitude of the relative velocity is maintained. The deflection angle $θ$ , is only a function of the impact parameter $b$ , when the intermolecular forces are given. The impact parameter is the distance of closest approach of the center of mass of the two particles, if the particle’s trajectory was not deflected by the collision. Since we are considering many particles, and want to know how they collide with each other, the differential cross section $\vartheta$ and the solid angle $d Ω$ are defined to account for all possible angles the plane of interaction may be as well as the deflection angle.

$d\Omega =\sin \left( \theta \right)d\theta d\phi \qquad \qquad( 18 )$

$\vartheta d\Omega =bdbd\phi \qquad \qquad( 19 )$

The azimuthal angle is $φ$ , which defines the plane in which a collision occurs. The collision function, ${{\Omega }_{i}}\left( {{f}_{i}} \right)$ , for only binary collisions is (Chapman and Cowling, 1970):

${{\Omega }_{i}}\left( {{f}_{i}} \right)=\iint{\left( {{{{f}'}}_{i}}{{{{f}'}}_{A}}-{{f}_{i}}{{f}_{A}} \right)\left| \mathbf{c}-{{\mathbf{c}}_{A}} \right|}\vartheta d\Omega {{d}^{3}}{{\mathbf{c}}_{A}}\qquad \qquad( 20 )$

The distribution functions after a collision are $f'$ and $f' A$ , and those before the collisions are $f$ and $f A$ . Since the collision function comes from the impact of particles that conserve mass, momentum and energy, the integration of the elementary collision invariants and the collision function are zero (Cercignani, 1975).

$\int{\psi \,{{\Omega }_{i}}\left( f \right){{d}^{3}}\mathbf{c}}=0\qquad \qquad( 21 )$

The elementary collision invariants are

$\psi =\left[ \begin{matrix} {{m}_{i}} \\ {{m}_{i}}\mathbf{c} \\ \frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}} \\ \end{matrix} \right]\qquad \qquad( 22 )$

The velocity, c, has three components. To gain a general transport equation, the Boltzmann equation can be multiplied by a transport variable, $ψ i$ , and integrating over ${{d}^{3}}\mathbf{c}$ yields:

$\int{\psi \left[ \frac{\partial {{f}_{i}}}{\partial t}+\mathbf{c}\cdot \nabla {{f}_{i}}+\mathbf{a}\cdot {{\nabla }_{\mathbf{c}}}{{f}_{i}} \right]{{d}^{3}}\mathbf{c}}=\int{\psi \,{{\Omega }_{i}}\left( f \right){{d}^{3}}\mathbf{c}}\qquad \qquad( 23 )$

If the transport variable is an invariant, than the right hand side of eq. (23) is zero. The first term in the derivative can be rewritten using the chain rule and Liebnitz integral rule.

$\int{\psi \frac{\partial {{f}_{i}}}{\partial t}{{d}^{3}}\mathbf{c}}=\frac{\partial }{\partial t}\int{\psi {{f}_{i}}{{d}^{3}}\mathbf{c}}-\int{{{f}_{i}}\frac{\partial {{\psi }_{i}}}{\partial t}{{d}^{3}}\mathbf{c}}\qquad \qquad( 24 )$

The second term can be rewritten as:

$\int{\psi {{\mathbf{c}}_{j}}\frac{\partial {{f}_{i}}}{\partial {{\mathbf{x}}_{j}}}{{d}^{3}}\mathbf{c}}=\frac{\partial }{\partial {{\mathbf{x}}_{j}}}\int{\psi {{\mathbf{c}}_{j}}{{f}_{i}}{{d}^{3}}\mathbf{c}}-\int{{{f}_{i}}{{\mathbf{c}}_{j}}\frac{\partial \psi }{\partial {{\mathbf{x}}_{j}}}{{d}^{3}}\mathbf{c}}\qquad \qquad( 25 )$

Note that the subscript i refers to a specific species and the subscript j refers to a direction. The independent variable is time, t, and x and c are both functions of time [ $\mathbf{x}=\mathbf{x}(t)$ and $\mathbf{c}=\mathbf{c}(t)$ ]; therefore, $\partial \mathbf{c}/\partial \mathbf{x}=0$ . Also, the acceleration is assumed to be independent of the particle velocity, which is valid when gravitational or electromagnetic forces are involved.

$\int{\psi {{\mathbf{a}}_{j}}\frac{\partial {{f}_{i}}}{\partial {{\mathbf{c}}_{j}}}{{d}^{3}}\mathbf{c}}=\frac{\partial }{\partial {{\mathbf{c}}_{j}}}\int{\psi {{\mathbf{a}}_{j}}{{f}_{i}}{{d}^{3}}\mathbf{c}}-\int{{{f}_{i}}{{\mathbf{a}}_{j}}\frac{\partial \psi }{\partial {{\mathbf{c}}_{j}}}{{d}^{3}}\mathbf{c}}\qquad \qquad( 26 )$

For simplicity, any function $K$ multiplied by $f i$ and integrated is written as:

$\int{K{{f}_{i}}{{d}^{3}}\mathbf{c}}=\bar{K}\qquad \qquad( 27 )$

Therefore, the Boltzmann equation can be rewritten as a general transport equation. This equation relates the microscale physics to macroscale physics.

$\frac{\partial }{\partial t}\overline{\psi }-\overline{\frac{\partial \psi }{\partial t}}+\nabla \cdot \left( \overline{\psi \mathbf{c}} \right)-\overline{\mathbf{c}\nabla \psi }+{{\nabla }_{\mathbf{c}}}\cdot \left( \overline{\psi \mathbf{a}} \right)-\overline{\mathbf{a}{{\nabla }_{\mathbf{c}}}{{\psi }_{i}}}=\overline{\psi \,{{\Omega }_{i}}}\qquad \qquad( 28 )$

To retain species continuity, $ψ = m i$ , and the mass of a particle does not change with time, location or velocity, $\frac{\partial {{m}_{i}}}{\partial t}=\frac{\partial {{m}_{i}}}{\partial \mathbf{x}}=\frac{\partial {{m}_{i}}}{\partial \mathbf{c}}=0$ . Therefore,

$\frac{\partial }{\partial t}\overline{{{m}_{i}}}+\nabla \cdot \left( \overline{{{m}_{i}}\mathbf{c}} \right)=\frac{\partial {{\rho }_{i}}}{\partial t}+\nabla \cdot \left( {{\rho }_{i}}{{\mathbf{V}}_{i}} \right)=0\qquad \qquad( 29 )$

To retain the momentum equation, the invariant may be set to the particle momentum, $\psi ={{m}_{i}}{{\mathbf{c}}_{j}}$ .

$\frac{\partial }{\partial t}\overline{{{m}_{i}}{{\mathbf{c}}_{j}}}-\overline{\frac{\partial {{m}_{i}}{{\mathbf{c}}_{j}}}{\partial t}}+\nabla \cdot \left( \overline{{{m}_{i}}{{\mathbf{c}}_{j}}\mathbf{c}} \right)=0\qquad \qquad( 30 )$

The fourth term drops out because c is not a function of x. The last two terms of the general transport equation drop out, because

${{\nabla }_{\mathbf{c}}}\cdot \left( \overline{{{m}_{i}}{{\mathbf{c}}_{j}}\mathbf{a}} \right)-\overline{\mathbf{a}{{\nabla }_{\mathbf{c}}}{{m}_{i}}{{\mathbf{c}}_{j}}}=\overline{{{m}_{i}}\left( \mathbf{a}-\mathbf{a} \right)}=0\qquad \qquad( 31 )$

The body force term comes from the second term in eq. (30), because ${{m}_{i}}\frac{\partial {{\mathbf{c}}_{j}}}{\partial t}={{m}_{i}}{{\mathbf{a}}_{j}}$ . The third term in eq. (30) may be the term of greatest interest, because it contains information about the microscopic and the bulk movements of particles.

$\overline{{{m}_{i}}{{\mathbf{c}}_{j}}\mathbf{c}}=\overline{\left( {{\mathbf{V}}_{j}}+{{\mathbf{w}}_{i,j}} \right)\left( \mathbf{V}+{{\mathbf{w}}_{i}} \right)}={{\rho }_{i}}{{\mathbf{V}}_{j}}\mathbf{V}+\overline{{{m}_{i}}{{\mathbf{w}}_{i,j}}{{\mathbf{w}}_{i}}}\qquad \qquad( 32 )$

The random velocity of species i in the j direction is represented by ${{\mathbf{w}}_{i,j}}$ . Now the momentum equation can be rewritten in terms of the average velocity.

$\sum\limits_{i}{\left[ \frac{\partial {{\rho }_{i}}{{\mathbf{V}}_{j}}}{\partial t}+\nabla \cdot \left( {{\rho }_{i}}{{\mathbf{V}}_{j}}\mathbf{V} \right)={{\rho }_{i}}{{\mathbf{X}}_{j}}-\nabla \cdot \left( {{\rho }_{i}}\overline{{{\mathbf{w}}_{i,j}}{{\mathbf{w}}_{i}}} \right) \right]}\qquad \qquad( 33 )$

Since this momentum equation is identical to the Navier-Stokes equation, the stress tensor is:

$\tau =-\sum\limits_{i}{{{\rho }_{i}}\overline{{{\mathbf{w}}_{i}}{{\mathbf{w}}_{i}}}}\qquad \qquad( 34 )$

The scalar gas pressure is defined as one-third of the trace of the stress tensor.

$p=-\sum\limits_{i}{\frac{1}{3}{{\rho }_{i}}\left( \overline{\mathbf{w}_{i,1}^{2}+\mathbf{w}_{i,2}^{2}+\mathbf{w}_{i,3}^{2}} \right)}\qquad \qquad( 35 )$

Pressure always has a positive value, as it does in this definition. To retain the remainder of the stress tensor, $τ'$ , the pressure can be subtracted from the total stress tensor.

$\begin{align} & {\tau }'=\sum\limits_{i}{-\rho \overline{_{i}{{\mathbf{w}}_{i}}{{\mathbf{w}}_{i}}}+\frac{1}{3}{{\rho }_{i}}\left( \overline{\mathbf{w}_{i,1}^{2}}+\overline{\mathbf{w}_{i,2}^{2}}+\overline{\mathbf{w}_{i,3}^{2}} \right)\mathbf{I}} \\ & =-\left[ \sum\limits_{i}{{{\rho }_{i}}\overline{{{\mathbf{w}}_{i}}{{\mathbf{w}}_{i}}}}-p\mathbf{I} \right]+\frac{1}{3}\left[ \left( \sum\limits_{i}{{{\rho }_{i}}\overline{\mathbf{w}_{i,1}^{2}}}-p \right)+\left( \sum\limits_{i}{{{\rho }_{i}}\overline{\mathbf{w}_{i,2}^{2}}}-p \right)+\left( \sum\limits_{i}{{{\rho }_{i}}\overline{\mathbf{w}_{i,3}^{2}}}-p \right) \right]\mathbf{I} \\ \end{align}\qquad \qquad( 36 )$

Reducing the stress tensor to be governed by Newton’s law of viscosity,

$\sum\limits_{i}{{{\rho }_{i}}\overline{{{\mathbf{w}}_{i}}{{\mathbf{w}}_{i}}}}-p\mathbf{I}=-\mu \left( \nabla \mathbf{V}+{{\left( \nabla \mathbf{V} \right)}^{T}} \right)\qquad \qquad( 37 )$

Now the total stress tensor can be written as:

$\tau =-\sum\limits_{i}{{{\rho }_{i}}\overline{{{\mathbf{w}}_{i}}{{\mathbf{w}}_{i}}}}=-p\mathbf{I}+\mu \left[ \nabla \mathbf{V}+{{\left( \nabla \mathbf{V} \right)}^{T}} \right]-\frac{2}{3}\mu \left( \nabla \cdot \mathbf{V} \right)\mathbf{I}\qquad \qquad( 38 )$

Summing the momentum equation over all the species, applying continuity and substituting the stress tensor, the momentum equation can be written as:

$\rho \frac{D\mathbf{V}}{Dt}=\nabla \cdot \left( \tau \right)+\rho \mathbf{X}\qquad \qquad( 39 )$

For the energy equation, the invariant may be set to the kinetic energy of a particle, $\psi =\frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}}$ .

$\sum\limits_{i}{\left[ \frac{\partial }{\partial t}\overline{\frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}}}-\overline{\frac{\partial \frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}}}{\partial t}}+\nabla \cdot \left( \overline{\frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}}\mathbf{c}} \right) \right]}=0\qquad \qquad( 40 )$

The fourth term drops out because $c 2$ is not a function of $x j$ . The last two terms of the general transport equation drop out, because

${{\nabla }_{\mathbf{c}}}\cdot \left( \overline{\frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}}\mathbf{a}} \right)-\overline{\mathbf{a}{{\nabla }_{\mathbf{c}}}\frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}}}=\overline{{{m}_{i}}\mathbf{c}\cdot \left( \mathbf{a}-\mathbf{a} \right)}=0\qquad \qquad( 41 )$

The energy added due to the body force comes from the second term in eq. (40), because $\overline{\frac{1}{2}{{m}_{i}}\frac{\partial {{\mathbf{c}}^{2}}}{\partial t}}=\overline{{{m}_{i}}\mathbf{c}\cdot \mathbf{a}}={{\rho }_{i}}{{\mathbf{V}}_{i}}\cdot \mathbf{X}$ . The first term can be rewritten in terms of the average velocity and the deviant velocity.

$\frac{\partial }{\partial t}\left( \overline{\frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}}} \right)=\frac{\partial }{\partial t}\left( \sum\limits_{i}{\frac{1}{2}{{\rho }_{i}}\left( \overline{\mathbf{w}_{i}^{2}}+{{\mathbf{V}}^{2}} \right)} \right)=\frac{\partial }{\partial t}\left( \rho \left( e+\frac{{{\mathbf{V}}^{2}}}{2} \right) \right)\qquad \qquad( 42 )$

The third term in equation (40) can also be expanded, resulting in several terms, such as advection, heat flux, and viscous dissipation terms.

$\begin{align} & \sum\limits_{i}{\nabla \cdot \left( \overline{\frac{1}{2}{{m}_{i}}{{\mathbf{c}}^{2}}\mathbf{c}} \right)}=\sum\limits_{i}{\nabla \cdot \left( \frac{1}{2}{{\rho }_{i}}\overline{\left( \mathbf{w}_{i}^{2}+2{{\mathbf{w}}_{i}}\mathbf{V}+{{\mathbf{V}}^{2}} \right)\left( \mathbf{V}+{{\mathbf{w}}_{i}} \right)} \right)} \\ & =\sum\limits_{i}{\nabla \cdot \left( \frac{1}{2}{{\rho }_{i}}\left( \overline{\mathbf{w}_{i}^{2}}+{{\mathbf{V}}^{2}} \right)\mathbf{V} \right)}+\sum\limits_{i}{\nabla \cdot \left( {{\rho }_{i}}\left( \frac{1}{2}\overline{\mathbf{w}_{i}^{2}{{\mathbf{w}}_{i}}}+\overline{{{\mathbf{w}}_{i}}\left( {{\mathbf{w}}_{i}}\cdot \mathbf{V} \right)} \right) \right)} \\ & =\nabla \cdot \left( \rho \left( e+\frac{{{\mathbf{V}}^{2}}}{2} \right)\mathbf{V} \right)+\sum\limits_{i}{\nabla \cdot \left( {{\rho }_{i}}\left( \frac{1}{2}\overline{\mathbf{w}_{i}^{2}{{\mathbf{w}}_{i}}}+\overline{{{\mathbf{w}}_{i}}\left( {{\mathbf{w}}_{i}}\cdot \mathbf{V} \right)} \right) \right)} \\ \end{align}\qquad \qquad( 43 )$

The last term includes the mechanical work of the stress tensor, as well as the heat flux components.

$\sum\limits_{i}{\frac{1}{2}{{\rho }_{i}}\overline{\mathbf{w}_{i}^{2}{{\mathbf{w}}_{i}}}}=\mathbf{{q}''}\qquad \qquad( 44 )$

$\sum\limits_{i}{{{\rho }_{i}}\overline{{{\mathbf{w}}_{i}}\left( {{\mathbf{w}}_{i}}\cdot \mathbf{V} \right)}}=\tau \cdot \mathbf{V}\qquad \qquad( 45 )$

Therefore, the energy equation can be rewritten in the standard form.

$\frac{\partial }{\partial t}\left( \rho \left( e+\frac{{{\mathbf{V}}^{2}}}{2} \right) \right)+\nabla \cdot \left( \rho \mathbf{V}\left( e+\frac{{{\mathbf{V}}^{2}}}{2} \right) \right)=-\nabla \cdot \mathbf{{q}''}+\nabla \cdot \left( \tau \cdot \mathbf{V} \right)+\rho \mathbf{V}\cdot \mathbf{X}\qquad \qquad( 46 )$

The bulk transport effects are derived from a statistical molecular model, to get equations governing conservation of mass, momentum and energy. Since only binary collisions are considered, and the rotational and vibrational energies of a molecule are neglected, the model is valid for dilute monatomic gases.

Another insight to the Boltzmann equation is looking at the equilibrium distribution function of a gas mixture at rest, known as the Maxwell-Boltzmann distribution. At equilibrium, the rate of change of the distribution function is zero:

$\frac{D{{f}_{i}}}{Dt}=0\qquad \qquad( 47 )$

In order for the rate of change of the distribution function to be zero, the collision function must also be zero, $Ω i = 0$ . Therefore:

${{{f}'}_{i}}{{{f}'}_{A}}={{f}_{i}}{{f}_{A}}\qquad \qquad( 48 )$

The solution to this function is the Maxwell-Boltzmann distribution function:

${{f}_{i}}\left( \mathbf{x},\mathbf{c},t \right)={{n}_{i}}{{\left( \frac{{{m}_{i}}}{2\pi {{k}_{B}}T} \right)}^{3/2}}\exp \left[ -\frac{{{m}_{i}}{{\left( \mathbf{c}-\mathbf{V} \right)}^{2}}}{2{{k}_{B}}T} \right]\qquad \qquad( 49 )$

This function describes how the particle velocities vary at each location.

Figure 2: Lattice Boltzmann model discretization and lattice structure for the (a) velocity vectors corresponding to

f i

of a node, (b) Lattice structure connecting nodes.

Modeling of the Boltzmann equation can be done using the Lattice Boltzmann model (McNamara and Zanetti, 1988). This model discretizes the Boltzmann equation into six velocity components (seven if a resting particle is included) at different locations on a mesh. A representation of the Lattice Boltzmann model is presented in Fig. 2. The mesh is constructed so that the grid-spacing of the nodes is equivalent to a velocity component multiplied by the time-step. The distribution function is broken into two parts: a streaming step and a collision step. The particles will stream along the grid “lattices” during the streaming step. During the collision step, a collision function is applied at the node. This collision function usually incorporates mass and momentum conservation, but generally the collisions should conserve mass, momentum, and energy.

References

Chapman, S. and Cowling, T.G., 1970, The Mathematical Theory of Non-uniform Gases, Cambridge University Press, Cambridge, UK.

Cercignani, C., 1975, Theory and Application of the Boltzmann Equation, Scottish Academic Press.

McNamara, G.R. and Zanetti, G., 1988, “Use of the Boltzmann Equation to Simulate Lattice-Gas Automata,” Physical Review Letters, Vol. 61, pp. 2332-2335.

Rice, J. 2006b, Personal Communication, University of Connecticut, Storrs, CT.

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Boltzmann Equation

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