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{latex}
\begin{aligneqnarray*}
Re \equiv \frac{\rho d_p |\overrightarrow{u_p}-\overrightarrow{u}|}{\mu}
\end{aligneqnarray*}
{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}
\[
uU_{initial}(y) =
\left\{
  \begin{array}{lr}
    tanh(32y-8) & : 0 \leq x \leq 0.5\\
    -tanh(32y-24) & : 0.5n < n \leq 1
  \end{array}
\right.
\]
{latex}

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{latex}$\tau${latex}

{latex}$U_0${latex}

{latex}$d_c${latex}

Thus: 

{latex}$d_c = d_p = 

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d 

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(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{aligneqnarray*}
&\tau = \frac{\rho_d d_d ^2}{18 \mu_g}(3) \\
\end{aligneqnarray*}
{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}

Expected Results

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