Question 8.13: Centrifugal Separator Process water of density 1000 kgm^–3 ......

A centrifugal separator consists of a fast rotating cylinder or bowl into which a liquid or slurry is fed containing a mixture of solid particles for separation. The liquid forms an annular body against the inner wall of the cylinder or bowl for which the clarified liquid overflows through orifices. The solids settle on the surface of the bowl as a result of centrifugal forces. The solids are pushed along by way of a screw conveyor. The basis of design is the same as the design of settling tanks. In this case, a high centrifugal force on the particles replaces the relatively weaker gravitational force to enhance the speed of separation. Stokes’ law (see Equation 8.1) is accordingly modified to

U_{t}={\frac{g\left(\rho_{p}-\rho\right)d_{p}^{2}}{18\mu}}           (8.1)

U_{t}={\frac{\omega^{2}r\left(\rho_{p}-\rho\right)d_{p}^{2}}{18\mu}}            (8.51)

The terminal velocity of a particle moving toward the bowl surface is equal to the change in radius with time. The time it takes for a particle initially at the surface liquid to reach the bowl surface can therefore be determined by integrating

\int_{r_{1}}^{r_{2}}{\frac{d r}{r}}={\frac{\omega^{2}\left(\rho_{p}-\rho\right)d_{p}^{2}}{18\mu}}\int_{0}^{t}d t            (8.52)

to give a rearranging of

t=\frac{18\mu\ln\left[\frac{r_{2}}{r_{1}}\right]}{\omega^{2}\left(\rho_{p}-\rho\right)d_{p}^{2}}= \frac{18\times10^{-3}\times\mathrm{ln}\left[\frac{0.25/2}{0.05/2}\right]}{\left(2\pi\times\frac{2000}{60}\right)^{2}\times(2800-1000)\times\left(30\times10^{-6}\right)^{2}} = 0.04 s             (8.53)

For the continuous operation of the centrifuge, the time for a particle to reach the bowl surface is equal to the residence time of the separator.