Terminal Velocity Most separators will have a given dimension stated by the terminal or maximum velocity achieved by particles. So in that manner it is the first expression to be obtained. Furthermore, for the case of maximum velocity , there will be no acceleration( a=0). Hence the equation becomes :

$m_pg-\rho_fV_pg-\frac{1}{2}C_DA_P\rho_f(u_{max})^2$

Given the volume and area of a sphere and the density expressed via Vp= corresponding volume of the particles, rp=volumetric radius

$d=2r_p=\left(\frac{6V_p}{\pi})^\frac{1}{3}\right)$

$A_p=A_{sphere}=4\pi r_{p}^{2}$

$A_p>A_{sphere}(non-sphereical)$

$\rho_p=\frac{m_p}{V_p}$

To consider the shape of the particle, surface area is used. An increase in it may cause an increase in drag.

$\frac{1}{2}C_D\pi r^2\rho_fu_{max}^2=V_fg(\rho_p-\rho_f)$

$\frac{1}{2}C_D\pi r^2\rho_fu_{max}^2=\frac{4}{3}\pi r^3g(\rho_p-\rho_f)$

$u_{max}^{2}=\frac{8rg(\rho_p-\rho_f)}{3\rho_fC_D}$

Leading to the expression for maximum velocity expressed as

$u_{max}=\sqrt{\frac{8rg(\rho_p-\rho_f)}{3\mu_f}}$

Which can also be expressed as : But due to the drag coeddiecent used ( based on stokes ), only applicable at low Re

$u_{max}=\frac{2r^2g(\rho_p-\rho_f)}{9\mu_f}$

Terminal velocity using the Stokes correction factor : Another way to express the terminal velocity is by the use of a correction factor . This cor- rection factor is a coefficient by which the flow in question would be assimilated to the one assumed by stokes following

$F_D=-3\pi d\mu_fu_f=-\frac{1}{8}\pi d^2\rho_f\vert u \vert C_D$

Thus yielding

$u_{max}\frac{2}{9}\frac{r^2(\rho_p-\rho_f)}{\mu_ff}$

Transient Velocity for Stokes Regime.

The transient velocity of such particle is time dependent. It will increase over the time until it reaches the final maximum velocity umax .What follows is the deriving of such expression :

$\sum F=m_pa=m_p\frac{du}{dt}=F_g-F_b-F_D$

$\sum F=m_pa=m_p\frac{du}{dt}=m_pg-\rho_fV_pg-\frac{1}{2}C_DA_P\rho_fu^2$

Using the same expressions as for solving for umax leads to

$\rho_pV_p\frac{du}{dt}=V_pg(\rho_p-\rho_f)-\frac{1}{2}C_D\pi r^2\rho_fu^2$

By substituting CD with the Reynolds dependent equation derived by Stokes expression and substituting this Reynolds with its definition yields

$\rho_pV_p\frac{du}{dt}=V_pg(\rho_p-\rho_f)-\frac{1}{2}\frac{24\mu_f}{2r\rho_fu}\pi r^2\rho_fu^2$

$\frac{2}{9}\frac{r^2\rho_p}{\mu_f}\frac{du}{dt}=\frac{2}{9}\frac{r^2g(\rho_p-\rho_f)}{mu_f}-u$

Which by introducing $\lambda$ , defined by someone as a particle relaxation time, which is assuming $\rho_f>>\rho_p ---> \delta\rho=\rho_p yeilds$

$\lambda\frac{2}{9}\frac{\rho_pr^2}{\mu_f}$

$\lambda\frac{du}{dt}=\lambda g-u$

Noticing that g = umax and integrating from zero to a time t in which a velocity u is present yields

$\lambda\frac{du}{dt}=u_{max} -u$

$u=u_{max}(1-exp\frac{-t}{\lambda})+Constant$

$u_{max}\frac{2}{9}\frac{r^2g(\rho_p-\rho_f)}{\mu_f}(1-exp(-f\frac{t}{\frac{2}{9}\frac{\rho_pr^2}{\mu_f}}))$

Notice that $\lambda$ is the coefficeint determining how fast terminal velocity is obtained.

Solving for the transient velocity profile for intermediate values of Re, equa- tions proposed by Putnam, Schiller-Naumann and similar have to be incor- porated and solved accordingly and incorporating them yields

$\rho_pV_p\frac{du}{dt}=V_pg(\rho_p-\rho_f)-/frac{1}{2}\pi r^2\rho_f(C_1u+C2u^k)$

which by introducing the constants a,b and c yields

$a=\frac{V_pg(\rho_p-\rho_f)}{\rho_pV_p}$

$b=\frac{\frac{1}{2}\pi r^2\rho_fC_1}{\rho_pV_p}$

$c=\frac{\frac{1}{2}\pi r^2 \rho_f C_2}{\rho_pV_p}$

$\frac{du}{dt}=a-bu-cu^k$

to solve with respect to a varying velocity , a solution for the integral is required. Hence a method for generally approximating the integral with respect to u is proposed.

$\int\frac{du}{a-bu-cu^k}$

This integral is to be solved in the open interval for which u varies , where u=z>0 .

Solving & approximation of the integral.Having proved that the approximation is reasoable, and processding with calculating the integral by susbituing the approximation into the integral, yields the following

$\int \frac{du}{a-bu-cu^k}$

This integral is trivial to solve and when computed yields

$\frac{A}{2(u_{top}-z}ln(\frac{u-2u_{top}+z}{w-z})+Bu+C$

â€‹where C is a constant of integration.