Question 8.9: Hydrocyclone Particle Separation Experimental tests involvi......

A hydrocyclone is a type of cyclone device used for the separation of particles suspended in a liquid. It is operated by feeding the liquid suspension tangentially into the top of the device to produce a centrifugal-type vortex action that allows the more dense and coarse particles to be separated in the underflow, while the smaller particles leave through the overflow. Hydrocyclones are typically used in mineral ore processing, drilling mud separation, and oil/water separation in offshore operations. With a low capital cost and inexpensive operational cost, they are effective at separating materials based on small differences in particle size.

Assuming that the particles are sufficiently far apart such that their interaction can be ignored and Stokes’ law applies (Equation 8.1), for which the terminal velocity of the mineral particles is

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

U_{t}={\frac{g(\rho_{p}-\rho)d_{p}^{2}}{18\mu}}={\frac{9.81\times(2800-1000)\times(5\times10^{-6})^{2}}{18\times0.001}}=2.45\times10^{-5}  {\mathrm{ms}}^{-1}                 (8.35)

For the mineral-in-oil suspension, the corresponding terminal velocity would have to be

U_{t2}=U_{t}{\frac{\dot{Q}_{2}}{\dot{Q}_{1}}}=2.45\times10^{-5}\times{\frac{0.04}{0.2}}=4.9\times10^{-6}  \mathrm{ms}^{-1}             (8.36)

Again, assuming Stokes’ law, the diameter of particles of the second mineral in oil would need to be

d_{p}={\sqrt{\frac{18\mathrm{\mu}U_{t}}{g(\mathrm{\rho}_{p}-\mathrm{\rho})}}} ={\sqrt{\frac{18\times0.01\times4.9\times10^{-6}}{9.81\times(1200-850)}}}=1.6\times10^{-5}\ {\mathrm{m}}           (8.37)