In this post, I want to share how to do preliminary sizing of hydrocyclones. In previous post, hydrocyclones are utilized for solid-liquid separations. The centrifugal force is produced by the motion of the liquid in this centrifugal device, which has a stationary wall. Like a gas cyclone, the gas cyclone operates on much the same principles. Hydrocyclones are inexpensive, reliable separators that work with particle sizes ranging from 4 to 500 micron. Figure below shows hydrocyclones typical proportion geometry.

Preliminary sizing of hydrocyclones can be done by using nomograph methods. The optimal arrangements and design for a certain application should be determined by consulting the expert manufacturers of hydrocyclone equipment.

Empirical equation below is used to estimate diameter of cyclone chamber.

Where:

d_{50} = the particle diameter for which the cyclone is 50% efficient, μm

*D _{c}* = diameter of cyclone chamber, cm

*μ* = liquid viscosity, cP

*L* = feed flow rate, l/min

*ρ _{L}* = density of liquid, g/cm

^{3}

*ρ _{g}* = density of gas, g/cm

^{3}

Equation above determines the hydrocyclone chamber diameter necessary to separate the so-called *d _{50}* particle diameter as a function of the slurry flow velocity and the physical characteristics of the liquid and solid. The separating efficiency for other particles is related to the

*d*diameter which is based on a formula:

_{50}Where:

ƞ = the efficiency of the cyclone in separating any particle of diameter *d*, in percent

d = the selected particle diameter, μm

The use of nomograph will be presented by using example below.

__Example__

Estimate the size of hydrocyclone needed to separate 90% of particles with a diameter greater than 20 μm, from 10 m^{3}/h of a dilute slurry.

Physical properties:

Solid density 2000 kg/m^{3}

Liquid density 1000 kg/m^{3}

Viscosity 1 cP

__Solution__

Flow rate = 10 m^{3}/h = 166.7 liter/min

*ρ _{S} – ρ_{L}* = 2 – 1 = 1 g/cm

^{3}

Based on figure below, for 90% removal of particles above 20 μm,

*d*_{50} = 14 μm

Based on figure below, for μ = 1 cP, *ρ _{S} – ρ_{L}* = 1 g/cm

^{3}, dan

*L*= 167 liter/min

__D _{c} __

__= 16 cm__Reference:

Coulson and Richardson’s Chemical Engineering Volume 6