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.