TECHNOLOGY FOR AIR POLLUTION
CONTROL – PART2
CYCLONES

Principle
•
The particles are removed by the application of a centrifugal
force. The polluted gas stream is forced into a vortex. the
motion of the gas exerts a centrifugal force on the particles, and
they get deposited on the inner surface of the cyclones
CYCLONES
CYCLONES (CONTD.)
Construction and Operation
The gas enters through the inlet, and is forced into a spiral.
•
•
At the bottom, the gas reverses direction and flows upwards.
To prevent particles in the incoming stream from
contaminating the clean gas, a vortex finder is provided to
separate them. the cleaned gas flows out through the vortex
finder.
CYCLONES (CONTD.)

Advantages of Cyclones
•
Cyclones have a lost capital cost
•
Reasonable high efficiency for specially designed cyclones.
•
They can be used under almost any operating condition.
•
Cyclones can be constructed of a wide variety of materials.
•

There are no moving parts, so there are no maintenance
requirements.
Disadvantages of Cyclones
•
They can be used for small particles
•
High pressure drops contribute to increased costs of operation.
PROBLEM

A cyclone with a flow rate of 150 m3/min has an
efficiency of 80%. Estimate the efficiency if the flow rate
is doubled.
SOLUTION
Step 1
Q1 = 150 m3/min
Q2 = 300 m3/min
Pt1 = 100% - 80% = 20%

Pt2/Pt1 = (Q1/Q2)0.5
Step 2
Final Efficiency = 1- Pt2
= 86%
DIMENSIONS OF A STANDARD CYCLONE
PROBLEM

Design of Cyclone
Body diameter = 0.75 m
Flow rate = 2.75 m3/s
ρp = 1600 kg/m3
ρg = 1.1 kg/m3
µ = 2.5*10-5 kg/m-s
High throughput
H = 0.8 * body diameter
W = 0.35 * body diameter
Lb = 1.7 * body diameter
Lc = 2.0 * body diameter
Calculate the diameter of particle with 50% efficiency
SOLUTION
Step 1:
Inlet Velocity = Q/A
= 17.46 m/s
 Step 2:
Number of effective turns
Ne = (Lb + Lc/2)/H
Ne = 3.375
 Step 3: Diameter of particle

dpc = 10 µm
FILTER CLOTH
FABRIC FILTERS
Fabric Filter
FABRIC FILTERS

Principle

The filters retain particles larger than the mesh size

Air and most of the smaller particles flow through. Some of the
smaller particles are retained due to interception and diffusion.

The retained particles cause a reduction in the mesh size.

The primary collection is on the layer of previously deposited
particles.
DESIGN OF FABRIC FILTERS
The equation for fabric filters is based on Darcy’s law for
flow through porous media.
 Fabric filtration can be represented by the following
equation:

S = Ke + Ksw
Where,
S = filter drag, N-min/m3 S = ∆P/V
Ke = extrapolated clean filter drag, N-min/m3
Ks = slope constant. Varies with the dust, gas and fabric, N-min/kg-m
W= Areal dust density = L V t
L = dust loading (g/m3), V = velocity (m/s)

Both Ke and Ks are determined empirically from pilot
tests.
Empirical Determination of Ke and Ks
PROBLEM

Estimate the values of Ke and Ks for the filter drag model:
Time (min)
5
10
15
20
25
30
Filter ∆P (Pa)
330
490
550
600
640
700
Limestone dust loading L = 1.00 g/m3
Fabric Area A = 1.00 m2
Air flow rate Q = 0.80 m3/min
SOLUTION
Step 1:
Calculate the air velocity
Air velocity = 0.80 (m3/min)/1.00 m2
= 0.80 m/min
 Step 2:

S = ∆P/V
412.5
612.5
687.5
750
800
875
W = LVt
4
8
12
16
20
24
Step 3:
Determine Ke and Ks graphically
Ke = 470 N-min/m3 Ks = 0.563 N-min/g-m

Fabric Filters
ΔP
Total pressure drop
Δ Pf
Pressure drop due to the fabric
Δ Pp
Pressure drop due to the particulate layer
Δ Ps
Pressure drop due to the bag house structure
ADVANTAGES OF FABRIC FILTERS

Very high collection efficiency

They can operate over a wide range of volumetric
flow rates

The pressure drops are reasonably low.

Fabric Filter houses are modular in design, and can be
pre-assembled at the factory
FABRIC FILTERS (CONTD.)

Disadvantages of Fabric Filters

Fabric Filters require a large floor area.

The fabric is damaged at high temperature.

Ordinary fabrics cannot handle corrosive gases.

Fabric Filters cannot handle moist gas streams

A fabric filtration unit is a potential fire hazard
Darcy’s equation
ΔPf
Pressure drop N/m2
ΔPp
Pressure drop N/m2
Df
Depth of filter in the direction of flow (m)
Dp
Depth of particulate layer in the direction of flow (m)
μ
Gas viscosity kg/m-s
V
superficial filtering velocity m/min
Kf, Kp
Permeability (filter & particulate layer m2)
60
Conversion factor δ/min
V = Q/A
Q
volumetric gas flow rate m3/min
A
cloth area m2
Dust Layer
L
t
ρL
Dust loading kg/m3
time of operation min
Bulk density of the particulate layer kg/m3
ΔP = ΔPf + ΔPp
Filter Drag S = ΔP/V
Areal dust density W = LVt
S= k1+k2W
PROBLEM

Estimate the net cloth area for a shaker bag house that
must filter 40,000 cfm of air with 10 grams of flour dust
per cubic foot of air. Also specify the number of
components to be used and calculate the total number
of bags required if each bag is 8 feet long and 0.5 feet in
diameter. The maximum filtering velocity for flour dust is
2.5 ft/min.
SOLUTION
Step 1:
Calculate total area and number of components required.
A = Q/V
 Step 2:
Calculate the area of each bag.
A = Π (d) l
 Step 3:
Calculate the total number of bags required.
Number of bags required = Total area / Area per bag
= 1270 bags
Number of compartments 4000 sq. ft. / compartment

ELECTROSTATIC PRECIPITATOR
Electrostatic Charging of Dust Particles
Cutaway of Electrostatic Precipitator
ELECTROSTATIC PRECIPITATOR

Principle





The particles in a polluted gas stream are charged by passing them through an
electric field.
The charged particles are led through collector plates
The collector plates carry charges opposite to that on the particles
The particles are attracted to these collector plates and are thus removed from
the gas steam
Construction and Operation of Electrostatic Precipitator



Charging Electrodes in the form of thin wires are placed in the path of the
influent gas.
The charging electrodes generate a strong electric field, which charges the
particles as they flow through it.
The collector plates get deposited with the particles. the particles are
occasionally removed either by rapping or by washing the collector plates.
DESIGN OF ELECTROSTATIC PRECIPITATORS

The efficiency of removal of particles by an Electrostatic
Precipitator is given by
η = fractional collection efficiency
w = drift velocity, m/min.
A = available collection area, m2
Q = volumetric flow rate m3/min
MIGRATION VELOCITY
Where,
q = charge (Columbus)
Ep = collection field intensity (volts/m)
r = particle radius (m)
μ = dynamic viscosity of gas (Pa-S)
c = Cunningham correction factor

Cunningham correction factor
where,
T = absolute temperature (°k)
dp = diameter of particle (μm)
PROBLEM

An ESP is designed to treat 50,000 m3/min with 97 %
efficiency. Assuming an effective drift velocity of 2.5
m/min, calculate the required plate area and the
number of plates. The plate size is 10 m by 5 m (height
by length).
SOLUTION
Step 1:
Efficiency of an Electrostatic Precipitator is given by

A =-[ (Q/w)*ln(1- η)]
A = 70,000 m2
Step 2:
Number of plates = total area/plate area
= 1400

ELECTROSTATIC PRECIPITATOR (CONTD.)

Advantages of Electrostatic Precipitators
 Electrostatic precipitators are capable very high efficiency,
generally of the order of 99.5-99.9%.
 Since the electrostatic precipitators act on the particles and not
on the air, they can handle higher loads with lower pressure
drops.
 They can operate at higher temperatures.
 The operating costs are generally low.

Disadvantages of Electrostatic Precipitators
 The initial capital costs are high.
 Although they can be designed for a variety of operating
conditions, they are not very flexible to changes in the operating
conditions, once installed.
 Particulate with high resistivity may go uncollected.
WET SCRUBBERS
Aeromix Wet Scrubber
WET SCRUBBERS
Floating Bed Wet Scrubber
WET SCRUBBERS

Principle

Wet scrubbers are used for removal of particles which have a
diameter of the order of 0.2 mm or higher.
 Wet scrubbers work by spraying a stream of fine liquid droplets on
the incoming stream.
 The droplets capture the particles
 The liquid is subsequently removed for treatment.

Construction and Operation

A wet scrubber consists of a rectangular or circular chamber in
which nozzles are mounted.
 The nozzles spray a stream of droplets on the incoming gas stream
 The droplets contact the particulate matter, and the particles get
sorbed.
 The droplet size has to be optimized.
WET SCRUBBERS (CONTD.)
o
Construction and Operation (contd.)

Smaller droplets provide better cleaning, but are more difficult
to remove from the cleaned stream.

The polluted spray is collected.

Particles are settled out or otherwise removed from the liquid.

The liquid is recycled.

Wet scrubbers are also used for the removal of gases from the
air streams.
SCRUBBER

Efficiency
where,
k = Scrubber coefficient (m3 of gas/ m3 of liquid)
R = Liquid-to-gas flow rate (QL/QG)
ψ = internal impaction parameter

Internal impaction parameter
where,
c = Cunningham correction factor
ρp = particle density (kg/m3)
Vg = speed of gas at throat (m/sec)
dp = diameter of particle (m)
dd = diameter of droplet (m)
μ = dynamic viscosity of gas, (Pa-S)
WET SCRUBBERS (CONTD.)

Advantages of Wet Scrubbers

Wet Scrubbers can handle incoming streams at high temperature, thus
removing the need for temperature control equipment.

Wet scrubbers can handle high particle loading.
 Loading fluctuations do not affect the removal efficiency.



They can handle explosive gases with little risk.
Gas adsorption and dust collection are handled in one unit.

Corrosive gases and dusts are neutralized.
Disadvantages of Wet Scrubbers


High potential for corrosive problems
Effluent scrubbing liquid poses a water pollution problem.
CYCLONE SPRAY CHAMBERS

These scrubbers combine a cyclone with a spray nozzle.

The added centrifugal force permits good separation of
the droplets, hence a smaller droplet size can be used.

Cyclone spray chambers provide up to 95% removal of
particles > 5 micron.
ORIFICE SCRUBBERS

The gas is impacted onto a layer of the scrubbing liquid.

The gas passes through the liquid, thus removing almost
all the particulate matter, and a large portion of the
probable gases.

After coming out of the liquid, the gas is passed through
baffles to remove the liquid droplets.
IMPINGEMENT SCRUBBERS

In Impingement scrubbers, the gas impacts a layer of
liquid/froth through a perforated tray.

Passing through this layer removes the particulate
matter.

The wet gas stream is then passed through a mist
collector.
VENTURI SCRUBBERS
VENTURI SCRUBBERS

The dirty gas is led in to the chamber at high inlet
velocities.

At the inlet throat, liquid at low pressure is added to the
gas stream

This increases the relative velocity between the gas and
the droplets, thus increasing the efficiency of removal.

Efficiencies of the range of 95% for particles larger than
0.2 mm have been obtained.
VENTURI SCRUBBER
Absolute Pressure Drop
Δp = pressure drop ( cm of water)
ug = gas velocity (cm/s)
Qt = liquid volume flow rate
Qg = gas volume flow rate
PROBLEM

Water is introduced into the throat of a venturi scrubber.
The air velocity through the scrubber is 550 fps and the
liquid to gas ratio is 8.5 gal/1000 actual ft3. Determine
the pressure drop?
SOLUTION
Step 1:
Absolute pressure drop is given by

Δp = 4.8
HYDROCARBON CONTROL
GENERAL METHODS FOR CONTROL OF
HYDROCARBON EMISSIONS

Incineration or after burning

Direct flame incineration

Thermal incineration

Catalytic incineration
DIRECT FLAME INCINERATION
CATALYTIC INCINERATION
THERMAL INCINERATOR
CATALYTIC INCINERATOR WITH HEAT RECOVERY
PROBLEM
 Calculate
the removal efficiency of a burner if the
concentration of HC was reduced from 1300 ppm to
100 ppm.
SOLUTION
Step 1:
Removal Efficiency of the burner

Efficiency = (initial concentration – final concentration)/ initial concentration
Efficiency of the burner = 92.3%
VOC INCINERATORS
VOC INCINERATORS

Principle

VOC incinerators thermally oxidize the effluent stream, in the
presence of excess air.

The complete oxidation of the VOC results in the formation of
carbon monoxide and water. The reaction proceeds as follows:
CxHy + ( x + y/4 ) O2 x CO2 + (y/2) H2O
 Operation
The most important parameters in the design and operation of an
incineration system are what are called the
' three T's ' Temperature, Turbulence, and residence Time.
VOC INCINERATORS (CONTD.)

Temperature


o
Timing

o
The reaction kinetics are very sensitive to temperature
The higher the temperature, the faster the reaction
A certain time has to be provided for the reaction to proceed
Turbulence


Turbulence promotes mixing between the VOC's and oxygen
Proper mixing helps the reaction to proceed to completion in
the given time.
VOC INCINERATORS (CONTD.)

The various methods for incineration are:

Elevated fires, for concentrated streams

Direct thermal oxidation, for dilute streams

Catalytic oxidation, for dilute streams.
PROBLEM

In a workshop a mixture of chemicals were released,
Benzene – 3000 ppm
Toulene – 1000 ppm
Methane – 2000 ppm
Calculate the lower exposure limit (LEL) of the mixture
LEL by volume for each chemical
Benzene – 1.4 %
Toulene – 1.27 %
Methane – 5.00%
SOLUTION
LEL of a mixture:
Xi - Volume of i component in the mixture
Xm - Volume of mixture
LELi - LEL of i component
LELmix = [3000/(6000*1.4) + 1000/(6000*1.27) + 2000/(6000*5)]-1
= 1.79%
GASES
AIR POLLUTION CONTROL FOR GASES

Adsorption Towers

Thermal Incernation

Catalytic Combustion
Air Correction Equipment for Gases and Vapors
ADSORPTION TOWERS

Principle

Adsorption towers use adsorbents to remove the impurities
from the gas stream.

The impurities bind either physically or chemically to the
adsorbing material.

The impurities can be recovered by regenerating the adsorbent.

Adsorption towers can remove low concentrations of impurities
from the flue gas stream.
ADSORPTION TOWERS (CONTD.)
 Construction
and Operation

Adsorption towers consist of cylinders packed with the adsorbent.
 The adsorbent is supported on a heavy screen
 Since adsorption is temperature dependent, the flue gas is
temperature conditioned.
 Vapor monitors are provided to detect for large concentrations in the
effluent. Large concentrations of the pollutant in the effluent indicate
that the adsorbent needs to be regenerated.

Advantages of Adsorption Towers




Very low concentrations of pollutants can be removed.
Energy consumption is low.
Do not need much maintenance.
Economically valuable material can be recovered during regeneration.