Pre-Analysis & Start-Up

In the Pre-Analysis & Start-Up step, we'll review the following:

• Theory for Fluid Phase
• Theory for Particle Phase
• Choosing the Cases
  

Pre-Analysis:

A particle laden ow is a multiphase ow where one phase is the uid and the other is dispersed particles. Governing equations for both phases are implemented in Fluent. To run a meaningful simulation, a review of the theory is necessary.

Fluid Phase:

In the simulations considered for this tutorial, the uid ow is a 2D perturbed periodic double shear layer as described in the rst section. The geometry is Lx = XXm, Ly = XXm, and the mesh size is chosen as x = Lx=nx in order to resolve the smallest vorticies. As a rule of thumb. One typically needs about 20 grid points across the shear layers, where the vorticies are going to develop. The boundary conditions are periodic in the x and y  directions. The fluid phase satisfies the Navier Stokes Equations:
- Type here: momentum equations
- Type here: continuity equation
where uf is the uid velocity, p the pressure, rhof the uid density and f is a momentum exchange term due to the presence of particles. When the particle volume fraction and the particle mass loading M = p=f are very small, it is legitimate to neglect the eects of the particles on the uid: f can be set to zero. This type of coupling is called one-way. In these simulations the uid phase is air, while the dispersed phase is constitued of about 400 glass beads of diameter a few dozens of micron. This satises both conditions 1 and M 1. 1
One way-coupling is legitimate here. See ANSYS documentation (link) for further details about the momentum exchange term.

Continuity Equation:

{latex}
\begin{align*}
\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \textbf{v})=0 \\
\Rightarrow \nabla \cdot \textbf{v} = 0 \\
\end{align*}
{latex}

Navier-Stokes Equation:

{latex}
\begin{eqnarray*}
\rho (\frac{d \textbf{v}}{dt}+\textbf{v} \cdot \nabla \textbf{v})=- \nabla p + \mu \nabla ^2 \textbf{v} + \textbf{f}
\end{eqnarray*}
{latex}

In this case, however, we have discrete particles in the flow. Since we are using a one-way coupled scheme, the fluid imposes a force on the particles, but not vice versa. By using the particle force balance equation listed below, the particle movement can be calculated by integrating the acceleration term.

Particle Force Balance:

{latex}
\begin{align*}
&\frac{d u_p}{dt} = F_D(\overrightarrow{u}-\overrightarrow{u_p})+\frac{\overrightarrow{g}(\rho_p-\rho)}{\rho_p}+\overrightarrow{F} \\
\end{align*}
{latex}

{latex}$\overrightarrow{F}${latex}

 is an additional acceleration (force per unit particle mass) term. 

{latex}$F_D(\overrightarrow{u}-\overrightarrow{u_p})${latex}

 is the drag force per unit particle mass.

{latex}$F_D${latex}

 can be calculated using the formula below:

{latex}
\begin{align*}
F_D = \frac{18\mu}{\rho_d d_p ^2} \frac{C_D Re}{24}
\end{align*}
{latex}

Here, 

{latex}$\overrightarrow{u}${latex}

 is the fluid phase velocity, 

{latex}$\overrightarrow{u_p}${latex}

 is the particle velocity, 

{latex}$\mu${latex}

 is the molecular viscosity of the fluid, 

{latex}$\rho${latex}

 is the fluid density, 

{latex}$\rho_p${latex}

 is the density of the particle, and 

{latex}$d_p${latex}

 is the particle diameter. Re is the relative Reynolds, number, which is defined as 

{latex}
\begin{eqnarray*}
Re \equiv \frac{\rho d_p |\overrightarrow{u_p}-\overrightarrow{u}|}{\mu}
\end{eqnarray*}
{latex}

Initial Field Velocity Function

The initial field setup of a jet is essentially two shear layers. The two shear layers are symmetrical to the centerline. A hyperbolic tangent velocity function is used to initialize velocity field of a shear layer. The hyperbolic tangent velocity functions are preferred over discontinuous piecewise functions because hyperbolic tangent functions are more physical. The initial field velocity function for the entire jet is given below:

{latex}
\[
U_{initial}(y) =
\left\{
  \begin{array}{lr}
    tanh(32y-8) & : 0 \leq x \leq 0.5\\
    -tanh(32y-24) & : 0.5n < n \leq 1
  \end{array}
\right.
\]
{latex}

{latex}
\begin{align*}
&V_{initial} = 0\\
&\text{Tip:}\\
&tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}\\
\end{align*}
{latex}

The function is plotted below (x velocity versus y) using simple MATLAB codes:

Stokes Number

The Stokes number is a dimensionless number that corresponds to the behavior of particles suspended in a fluid flow. The Stokes number is defined as below:

{latex}
\begin{eqnarray*}
Stk = \frac{\tau U_0}{d_c}  (1)
\end{eqnarray*}
{latex}

{latex}$\tau${latex}

 is the relaxation time of the particle, 

{latex}$U_0${latex}

 is the fluid velocity well away from the particle, and 

{latex}$d_c${latex}

 is the characteristic length scale of the obstacle, in this case it is the diameter of the particle.

Thus: 

{latex}$d_c = d_p = d (2)${latex}

In the case of Stokes flow, where the Reynolds number is sufficiently small the relaxation time can be expressed using the equation below:

{latex}
\begin{eqnarray*}
\tau = \frac{\rho_d d_d ^2}{18 \mu_g}(3) \\
\end{eqnarray*}
{latex}

Plug (2) and (3) into (1), we get the formula for calculating the Stokes Number in our case below:

{latex}
\begin{eqnarray*}
\boxed{Stk = \frac{\rho_d \cdot d U_0}{18\mu_g}}\\
\end{eqnarray*}
{latex}

Here, 

{latex}$\rho_d${latex}

 is the density of the particle, 

{latex}$\mu_g${latex}

 is the dynamic viscosity of the fluid, in this case the dynamics viscosity of the gas.

Expected Results

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