cyclone separator 3

7/23/2019 Cyclone Separator 3

1/4

Lecture 3

Design of cyclones


2/4

DESIGN OF CYCLONESCyclone separators utilizes a centrifugal forces generated by a spinning gas stream to

separate the particulate matters from the carrier gas. The centrifugal force on particles in a

spinning gas stream is much greater than gravity; therefore cyclones are effective in the removal

of much smaller particles than gravitational settling chambers, and require much less space to

handle the same gas volumes.

In operation, the particle-laden gas upon entering the cyclone cylinder receives a rotating

motion. The vortex so formed develops a centrifugal force, which acts to particle radially

towards the wall. The gas spirals downward to the bottom of the cone, and at the bottom the gas

flow reveres to form an inner vortex which leaves through the outlet pipe [1].

Theory

In a cyclone, the inertial separating force is the radial component of the simple centrifugal

force and is a function of the tangential velocity. The centrifugal force can be expressed by Fc

r

mvF ec

2

(2.3.1)

Where, m=mass of the particle, ve=tangential velocity of the particle at radius r, and

r=radius of rotation. The separation factor S is given by

gr

vS e2

(2.3.2)

The separation factor varies from 5 in large, low velocity units to 2500 in small, high

pressure units. Higher the separation factor better is the performance of the cyclone.

In the cyclone, the gas, in addition to moving in a circular path, also moves radially

inwards between the inlet on the periphery and the exit on the axis. Since the tangential

velocities of the particle and the gas are the same, the relative velocity between the gas and

particle is simply equal to the radial velocity of the gas. This result in a drag force on the particletowards the centre, and the equilibrium radius of rotation of the particle can be obtained by

balancing the radial drag force and the centrifugal force:

r

vdvd gpprpg

23

63

(2.3.3)


3/4

Where, dp=particle diameter, and vr=radial velocity of the gas at radius r. Arranging the

above equation, for vr

r

vdv

g

gpp

r

18

23

(2.3.4)

The tangential velocity of the particle in the vortex has been found experimentally to be

inversely proportional to the radius of rotation according to equation,

constannrv (2.3.5)

Where, n is the exponent and dimensionless. For an ideal gas n=1. The real values

observed are between 0.5 to 1, depending upon the radius of the cyclone body and gas

temperature. can be related to the tangential velocity at the inlet to the cyclone as

n

ir

Dvv

2

(2.3.6)

Where, D=diameter of the cyclone. may be taken as the velocity of the gas through

the inlet pipe, i.e.,

i

iA

Qv

(2.3.7)

Where, Q=gas volumetric flow rate and Ai=cross-sectional area of the inlet. Therefore,

n

i r

D

A

Qv

2

(2.3.8)

n

i

n

g

gpp

rr

D

A

Q

r

dv

22

12

3

218

(2.3.9)

The most satisfactory expression for cyclone performance is still the empirical one.

Lapple correlated collection efficiency in terms of the cut size dpe which is the size of those

particle that are collected with 50% efficiency. Particle larger than dpe

will have collection

efficiency greater than 50% while the smaller particle will be collected with lesser efficiency.

The cut size is given by:

gpie

g

pevN

bd

2

9

(2.3.10)

v iv

iv


4/4

Where, b=inlet width, vi=gas inlet velocity andNe=effective number of turns a gas makes in

traversing the cyclone (5 to 10 in most cases).

Pressure drop:The pressure drop may be estimated according to the following equation,

2

2

2 e

ig

D

abvK

P

(2.3.11)

Where, K=a constant, which averages 13 and ranges from 7.5 to 18.4, =pressure

drop, a, b and De=cyclone dimensions, vi=inlet gas velocity and =gas density.

Problem 2.3.1: A conventional cyclone with diameter 0.5 m handles 4.0 m3/s of standard air

(g=1.8110-5

kg/m-s and gbeing negligible w.r.t p) carrying particles with a density of 2500

kg/m3. For Ne=6, inlet width (b)=0.25 m, inlet height (a)=0.5 m, determine the cut size of particle

diameter.

Solution: Given

b=0.25, D=0.250.5=0.1

a=0.5, D=0.50.5=0.25

p=2500 kg/m3

g=1.8110-5

kg/m-s

Q=4 m3/s

i

Q 4v =160 m/sa b 0.10.25

g

pe

e i p g

9 bd

2 N v

4

pe

9 1.81 0.25d 5.195 10 m

2 6 160 2500

REFERENCES

http://www.epconindia.com/air-pollution-control-equipment.html

P

g

cyclone separator 3

Documents