Cyclone Separator Challenge
Aim:- To perform analysis on cyclone separator and calculate the separation efficiency and pressure drop.
Objective:-
- To write a few words about any four empirical models used to calculate the cyclone separator efficiency.
- To perform an analysis on a given cyclone separator model by varying the particle diameter from 1 μm to 5 μm and calculate the separation efficiency in each case. Discuss the results. [Use both the velocity's as 3m/sec.]
- To perform an analysis on a given cyclone separator model by varying the particle velocity from 1 m/sec to 5 m/sec and calculate the separation efficiency and pressure drop in each case. Discuss the results. [Use particle diameter size as 5 μm]
Boundary Conditions:-
Inlet and wall:- Reflect
Outlet(Top):- Escape
Outlet(Bottom/dustbin) - Trap
Introduction
Cyclone separators are separation devices that use the principle of inertia to remove particulate matter from flue gases. Cyclone separators work much like a centrifuge, but with a continuous feed of dirty air. In a cyclone separator, dirty flue gas is fed into a chamber. The inside of the chamber creates a spiral vortex, The lighter components of this gas have less inertia, so it is easier for them to be influenced by the vortex and travel up it. Contrarily, larger components of particulate matter have more inertia and are not as easily influenced by the vortex.
Cyclone separators are generally able to remove somewhere between 50-99% of all particulate matter in flue gas. How well the cyclone separators are actually able to remove this matter depends largely on particle size. If there is a large amount of lighter particulate matter, less of these particles are able to be separated out. Because of this, cyclone separators work best on flue gases that contain large amounts of big particulate matter.
Empirical models used to calculate the cyclone separator efficiency
Iozia and Leith Model
Iozia and Leith logistic model is a modified version of Barth Model, which is developed based on force balance. The model assumes that a particle carried by the vortex endures the influence of two forces: a centrifugal force, Z and a flow resistance, W. The collection efficiency `eta`i of particle diameter dpi can be calculated from
`eta_i=1/(1+(d_(pc)/d_(p i))^beta`
`beta` is an expression for slope parameter derived based on the statistical analysis of experimental data of a cyclone with D =0.25 m given as
`beta=0.62-0.87ln(d_(pc)/100)+5.21ln((ab)/D^2)+1.05[ln((ab)/D^2)]^2`
Li and Wang Model
The Li and Wang model includes particle bounce or re-entrainment and turbulent diffusion at the cyclone wall. A two dimensional analytical expression of particle distribution in the cyclone is obtained. Li and Wang model was developed based on the following assumptions:
• The radial particle velocity and the radial concentration profile are not constant for uncollected particles within the cyclone.
• Boundary conditions with the consideration of turbulent diffusion coefficient and particle bounce re entrainment on the cyclone wall.
The resultant expression of the collection efficiency for particle of any size is given as
`eta_i=1-exp{-lambda theta_1}`
`theta_1=(2pi(S+L))/a`
Koch and Licht Model
Koch and Licht collection theory recognized the inherently turbulent nature of cyclones and the distribution of gas residence times within the cyclone. Koch and Licht described particle motion in the entry and collection regions with the additional following assumptions:
The tangential velocity of a particle is equal to the tangential velocity of the gas flow, i.e. there is no slip in the
tangential direction between the particle and the gas.
• The tangential velocity is related to the radius of cyclone by: uRn = constant.
A force balance and an equation on the particles collection yields the grade efficiency `eta_i`
`eta_i=1-exp{-2[(Gtau_iQ)/D^3(n+1)]^(0.5/(n+1))}`
Where,
`G=(8K_c)/(K_a^2 K_b^2)`
`n=1-{1-(12D)^0.14/2.5}{(T+460)/530}^0.3`
`tau_i=(rho_p d_(p i)^2)/(18mu)`
Lapple Model
Lapple modelwas developed based on force balance without considering the flow resistance. Lapple assumed that a particle entering the cyclone is evenly distributed across the inlet opening. The particle that travels from inlet half width to the wall in the cyclone is collected with 50% efficiency. The semi empirical relationship developed by Lapple to calculate a 50% cut diameter, dpc, is
`d_(pc)=[(9mub)/(2piN_enu_i(rho_p-rho_g))]^(1/2)`
Where `N_e`is the number of revolution
`N_e=1/a[h+(H-h)/2]`
The efficiency of collection of any size of particle is given by
`eta_i=1/(1+((d_(pc))/(dp i))^2`
Geometry
Import the given cyclone seperator in Ansys fluent
Volume extract
For the CFD analysis we created fluid domain by the tool Volume Extract inside ANSYS Spaceclaim and unchek the solid body supress for physics to hide the solid body .
Meshing
Setup
- pressure-based Steady state solver
- Velocity formulation - Absolute
- Gravity Acceleration
- X-direction = 0 m/s^2
- Y- direction = -9.81 m/s^2
- For Turbulance model follow the RNG K-epsilon method with Swirl dominated flow
-
Discrete phase model :
A discrete phase model should be set when the flow of a discrete phase (particles) with a continuum is modelled.
- Interaction with continous phase enable.
- Update DPM sources Every flow iteration : DPM iteration interval = 10
Injection Properties :
- Injection type = Surface
- Surface = Inlet
- Particle type = Inert
- Material = Anthracite
- Dimeter distribution = Uniform
- part-I
- case-1
- Variable
- Velocity in X-direction = 3 m/s
- Diameter = 1e-06 m
- Total flow rate = 1e-20 Kg/s
- Variable
- case-2
- Variable
- Velocity in X-direction = 3 m/s
- Diameter = 3e-06 m
- Total flow rate = 1e-20 Kg/s
- case-3
- Variable
- Velocity in X-direction = 3 m/s
- Diameter = 5e-06 m
- Total flow rate = 1e-20 Kg/s
- part-II
- case-1
- Variable
- Velocity in X-direction = 1 m/s
- Diameter = 5e-06 m
- Total flow rate = 1e-20 Kg/s
- case-2
- Variable
- Velocity in X-direction = 3 m/s
- Diameter = 5e-06 m
- Total flow rate = 1e-20 Kg/s
- case-3
- Variable
- Velocity in X-direction = 5 m/s
- Diameter = 5e-06 m
- Total flow rate = 1e-20 Kg/s
- Variable
-
Solution Method :
- Pressure velocity coupling - Scheme - SIMPLE
- Spatial discretization
- Gradient = Least squares cell based
- Pressure = Second order
- Momentum = First order upwind
- Turbulent kinetic energy = First order upwind
- Turbulent dissipation rate =First order upwind
Initialization :
- Hybrid initialization
Run calculation
- Number of iteration = 600
-
Boundary Conditions:-
Inlet and wall:- Reflect
Outlet(Top):- Escape
Outlet(Bottom/dustbin) - Trap
Part-1
Residual plot
case-1
Case-2
Case-3
Anthracite particle time snap shot
Case-1
Case-2
Case-3
Vortex core snap shot
Case-1
Case-2
Case-3
Part-2
Residual plot
Case-1
Case-2
Case-3
Anthracite particle time snap shot
Case-1
Case-2
Case-3
Vortex core snap shot
Case-1
Case-2
Case-3
Conclusion
Note : -ve sign in the mass flow from outlet-1 & outlet-2 means that the mass is release from the body to the atmosphere or outside.
1 When particles size increases number of trapped particle increases.
2 For part1 efficiency is more for the 5 micrometer particle size with respect to 1 micrometer particle size, that means when particle size increases seperator Efficiency increases.
3. From the above Anthracite particle time image we can say that Anthracite particle time is lesser for greater particle size.
4. From the above table we can see that mass flow rate remains constant when particle size increased.
5. From above table we can say that pressure does not varies when particle size increses .
6. From the above Anthracite particle time image we can say that Anthracite particle time is Smaller for same particle size when velocity increases
7. From the above table we can say that when velocity increases for the same particle size mass flow rate and pressure increased.
Seperator Efficiency and pressure drop for part1
Case-1
Seperator efficiency= Trapped/(Number Tracked-incomplete)
=205/(435-2)
= 0.47
Pressure drop=Inlet - outlet-2
= 17.7496-2.3110
= 15.4386 `N/m^2`
Case-2
Seperator efficiency= Trapped/(Number Tracked-incomplete)
= 248/435
= 0.57
Pressure drop=Inlet - outlet-2
= 17.7496-2.3110
= 15.4386 `N/m^2`
Case-3
Seperator efficiency= Trapped/(Number Tracked-incomplete)
= 317/435
= 0.72
Pressure drop=Inlet - outlet-2
= 17.7496-2.3110
= 15.4386 `N/m^2`
Seperator Efficiency and pressure drop for part2
Case-1
Seperator efficiency= Trapped/(Number Tracked-incomplete)
= 110/(435-110)
= 0.33
Pressure drop=Inlet - outlet-2
= 1.7517-0.2731
= 1.4786 `N/m^2`
Case-2
Seperator efficiency= Trapped/(Number Tracked-incomplete)
= 317/435
= 0.72
Pressure drop=Inlet - outlet-2
= 17.7496-2.3110
= 15.4386 `N/m^2`
Case-3
Seperator efficiency= Trapped/(Number Tracked-incomplete)
= 396/435
= 0.91
Pressure drop=Inlet - outlet-2
= 51.4403-6.3191
= 45.1212 `N/m^2`