1. AIR POLLUTION

2. OBJECTIVES Ability to define air pollution
Ability to describe classification of pollutant and particulate.
Ability to describe the effect of pollutant to health, plants and material.
Ability to explain stability of atmosphere including plume types.
Air quality control and treatment of emission products

3. DEFINITION
Air pollution is the presence in the outdoor atmosphere of one or more air contaminants (i.e., dust, fumes, gas, mist, odour, smoke, or vapour) in sufficient quantities, of such characteristics, and of such duration as to be or to threaten to be injurious to human, plant, or animal life or to property, or which reasonably interferes with the comfortable enjoyment of life or property’

4. CLSSIFICATION OF POLLUTANT
Pollutant can be classified into THREE that are:
1) Origin
Primary pollutants – such as SOx, NOx and hydrocarbon (HC) are those emitted directly to the atmosphere and were found in the form in which they were emitted.
Secondary Pollutants – such as O3 and peroxyacethyl nitrate (PAN) are those formed in the atmosphere as the result of chemical reactions among the primary pollutant and chemical species present in the atmosphere.

5. 2) Chemical Composition
Organic compounds – contain carbon & hydrogen, and may also contain element such as O, N , P and S.
Inorganic compounds – found in contaminated atmosphere include CO, CO2, carbonates, SOx, NOx,O3, HCl.

6. 3) State of Matter
Particulate pollutants – finely divided solids and liquids, include dust, fumes, smoke, fly ash, mist and spray.Under proper properties, particulate pollutant will settle.
Gaseous pollutants – formless fluids that completely occupy the space into which they are released, behave more as air & do not settle down. Including vapors of substance that are liquid or solid at normal Pressure & T. e.g: COx, SOx, NOx, HC and oxidants.

7. PARTICULATES
Particles can be classified from their mode of formation as dust, smoke, fumes, fly ash, mist, or spray.
The sizes range from 1000 µm – 0.01 µm.
The particles sizes from 100 µm – 0.01 µm are the major interest in air pollution studies because within these sizes the particulates can easily settle in lower respiratory tract (LRT).

8. small, solid particles created by the breakup of larger masses through processes such as crushing, grinding or blasting, may come directly from the processing or handling of materials such as coal, cement or grains. The size range from 1.0 µm to µm.
Dust
fine, solid particles resulting from the incomplete combustion of organic particles such as coal, wood or tobacco consists mainly of carbon and other combustible materials. The size range from 0.5 µm to 1 µm.
Smoke
fine, solid particles (often metallic oxides such as zinc and lead oxides) formed by the condensation of vapors of solid materials. Fumes are from sublimation, distillation, calcination or molten metal processes. The size range from 0.03 µm to 0.3 µm.
Fumes

9. finely divided solid particle, noncombustible particles contained in flue gases arising from combustion of coal and it is released when the organic portion of coal is burned. The size range from 1.0 µm to 1000 µm.
Fly ash
Mist
liquid particles or droplet formed by condensation of vapor, dispersion of a liquid (e.g. in foaming & splashing) or enactment of chemical reaction (e.g. formation of H2SO4). The sizes of mist <10 µm.
Spray
liquid particles formed by atomization of parent liquids (e.g. pesticides & herbicides). The size range from 10 µm to 1000 µm.

10. Air Quality Measurements
Particulates Measurments
Mini Volume Portable Sampler
High Volume Sampler

11. Air Quality Measurements
The filter is weighted before and after and the difference is the particulates collected
the air flow is measured by a small flow meter, usually calibrated every 24 hours because its get dirty during 24 hours of operations.
Less air goes through the filter during later part of the test than the beginning. Therefore air flow has to measure at both points , starting and ending points.
since the later part has different air flow than the beginning, air flow value it’s the average value of the start and end of the test.

12. Air Quality Measurements
Ex1: A clean filter is found to weigh 10 grams. After hours in hi-vol , the filter plus dust weighs 10.1 grams. The air glow at the start and end of the test are 60 & 40 m3/min respectively. What is the dust particulate concentration?

13. Air Quality Measurements
Gaseous measurements
The concentration of gasses can be either part per million (ppm) on a volume to volume basis or micrograms per cubic meter.
ppm means 1 volume of pollutants per 106 volume of Air. Converting µg/m3 needs to know the molecular weigh of the gas. At standard condition 00C and 1 Atmosphere pressure, one mole of gas occupies 22.4 L while at 250C and 1 Atmosphere pressure, one mole of gas occupies 24.5 L

14. Air Quality Measurements
Ex 2: A stack gas contains carbon monoxide (CO) as a concentration of 10% by volume. What is the concentration of (CO) in microgram per cubic meter. If CO MW is 28 g/mol, assume 250C and 1 Atmosphere pressure.

15. UNITS OF MEASUREMENT
There are 3 basic units used in reporting air pollution data that are micrograms per cubic meter (µg/m3), parts per million (ppm) and the micron (µ) or preferably known as micrometer (µm).
Micrograms per cubic meter and parts per million are unit of measurement for concentration and they are used to indicate the concentration of gaseous pollutant.
The µm is used to report the particle size.
Formerly concentration of gaseous pollutants were usually reported in parts per million (ppm), parts per hundred million (pphm), or parts per billion (ppb) by volume.
Thus, designations in µg/m3 may be followed by equivalent concentration on a ppm basis - e.g. 80 µg/m3 (0.03 ppm) of sulfur dioxide.

16. µg/m3 = ppmx 10-6 xGMW x103 L/m3 x106 µg/g L/mol
For gases, ppm can be converted to µg/m3 by using the following formula:
µg/m3 = ppmx 10-6 xGMW x103 L/m3 x106 µg/g
L/mol
GMW = gram molecular weight of gas
‘L/mol’ term is influenced by the temperature (T) and pressure (P) of the gas. According to Avogadro’s Law, 1 mole of any gas occupies the same volume as 1 mole of any gas at the same T and P.
Therefore, at 273 K (0OC) and 1 atm pressure (760 mmHg /101.3 kPa), standard conditions for many chemical reactions, the volume is 22.4 L/mol.

17. To convert to L/mol at other conditions, the following formula can be used:
V1P1 = V2P2
T T2
where V1, P1 and T1 is relate to the standard condition V2, P2 and T2 is relate to the actual condition that is being considered.

18. EXAMPLE :
Calculate the volume occupied by 4 mol of gas at 21.1OC and 760 mmHg.
SOLUTION:
V1P1 = V2P2
T T2
V1 = 22.4 L/mol P1 = 760 mmHg T1 = 273K
V2 = ? P2 = 760 mmHg
T2 = = 294.1K
4mol x 22.4 L/mol x 760mmHg = V2 x 760mmHg
273K K
V2 = L

19. DO IT YOURSELF
Determine the volume of 3 mol of stack gas at 1400 mmHg and 1000OC.
Ans: L

20. EXAMPLE :
The NO2 content of a sample of stack gas measured at 950OC at 2 atm pressure was 9 ppm. Determine the NO2 concentration in µg/m3 and mg/m3. NO2 weight 46 g/mol.
SOLUTION:
Step 1 – Find V2
V1P1 = V2P2
T T2
22.4L/mol x 1 atm = V2 x 2 atm
273K (273K + 950)

21. V2 = L/mol
Step 2 – Substitute V2 in Equation 1
µg/m3 = ppm x 10-6 x GMW x 103 L/m3
L/mol
µg/m3 = 9 x 10-6 x 46 g/mol x103 L/m3 x 106 µg/g
50.17 L/mol
= µg/m3 = mg/m3

22. DO IT YOURSELF
Gas from thermal pool has an SO2 content of 80 µg/m3 at kPa and 50OC. Calculate the SO2 concentration in ppm.

23. EXAMPLE :
What volume would one mole of an ideal gas occupy at 25.00C and kPa?.
SOLUTION:

24. Determine the volume of 6 mol of gas at 370C and 700 mmHg.
EXAMPLE :
Determine the volume of 6 mol of gas at 370C and 700 mmHg.
SOLUTION:

25. Convert 7.5 ppm of 64 g/mol SO2 to µg/m3 at 80OC and 110.5 kPa.
EXAMPLE :
Convert 7.5 ppm of 64 g/mol SO2 to µg/m3 at 80OC and kPa.
SOLUTION:
Step 1: Find V2

26. Step 2: Calculate SO2 concentration in μg/m3

27. Effects of Acidic Rain
-The potential effects of acid rain are relates to the acidity on aquatic life, damage to crops and forests, and damage to building materials.
-Lower pH values may affect fish directly by interfering with reproductive cycles or by releasing otherwise insoluble aluminium (Al), which is toxic.
Acid rain also leaches calcium (Ca) and magnesium (Mg) from the soil thus lower the molar ratio of Ca to Al which in turn, favours the uptake of Al by fine roots, that ultimately leads to their deterioration.

28. AIR Quality Control Treatment of Emissions

29. AIR Quality Control Selection of treatment devices will be depend on :
1- size of Air pollutant
2- Type of Air pollutant, such as SO2 can clean by water spray but the result is corrosion problem.
Therefore, are divided into different types for Air pollutant control process.

30. AIR Quality Control Where:Xo X1 X2
Where:
Xo : Amount of pollutant entering the device per unit time (kg/s)
X1 : Amount of pollutant collected by treatment device per unit time (kg/s)
X2 : Escaped particles (kg/s)

31. AIR Quality Control Where: R: Treatment device removal efficiency
Concentration of emission will be equal to the mass per volume.
Principles of mass balance will be valid for this process

32. AIR Quality Control
EX: An air pollution control device is to remove a particulate that is being emitted at a concentration of µg/m3 at an air flow rate of 180 m3/s. the device removes 0.48 ton per day. What are the emission concentration and the collection recovery?

33. control of Primary Particulates

34. Introduction
The control of primary particles is a major part of air pollution control engineering.
Although primary particles are generally larger than secondary particles , but many of them are small enough to be respirable and are thus of health concern.
If possible the collected particles are recycled to somewhere in the in the process that generates them.

35. Wall collection devices
Three types of control devices will be consider are, gravity settlers
Cyclone separators , and
Electrostatic precipitators
All function by driving the particles to a solid wall , where they adhere to each other to form agglomerates that can be removed from the collection device and disposed of.

36. gravity settlers
Easy to construct
Requires littlie maintenance
Treating very dirty gases in some industries
Contaminated gas passes slowly, allowing time for the particles to settle by gravity to the bottom.

38. Either the fluid going through is totally unmixed (block or plug flow model) or
The total mixing in the entire device or in the entire cross section, perpendicular to the flow (backmixed or mixed model)
For both cases the average horizontal gas velocity in the chamber is :

39. For the block flow model, assume :
1- the horizontal velocity of the gas in the chamber is equal to Vavg. Everywhere in the chamber
2- the horizontal component of the velocity of the particles in the gas is always equal to Vavg
3- the vertical component of the velocity of the particles is equal to their terminal settling velocity due to gravity Vt
4- if a particle settles to the floor , it stays there and is not re-entrained.

40. If the distance is greater than or equal to h, then it will reach the floor of the chamber and be captured.
For the same size particles , what will happen ?

41. For mixed flow model, assume:
1- the gas flow is totally mixed in the z direction but not in the x direction.
Most real gas flows are turbulent, leading to internal mixing in process equipment.
This because mixing in x direction moves particles up and downstream with the little effect on collection efficiency.
Whereas mixing in the z direction leads to a decrease in the collection efficiency.

42. Consider a section of the settler with the length dx.
The fraction of the particles in that section that reach the floor will equal the vertical distance an average particle fall due to gravity divided by the height of the section
The change in concentration passing this section is :

43. The time the average particle takes to pass through this section is :
Integrate from the inlet (x=0) to outlet (x=L) finding

44. Or
Substitute Vt from stake's law
Comparing this result with the plug or block assumption

45. Example 1
Compute the efficiency-diameter relation for a gravity settler that has
H=2m, L= 10 m and Vavg= 1 m/s
Assuming stake's law for both models
Particles density 2000 kg/m3 ,
Viscosity 1.8x10-5 Kg/m.s
Particle diameter in micron
1,10,30,50,57.45,80,100,120

46. EX: Design a settling chamber to collect particles 50 µm in diameter and 2000 Kg/m3 in density from an air stream with a volumetric flow of 1.5 m3 /s. the chamber is to be 2.0 m in width and 2.0 m in height. Average velocity 2 m/s, Viscosity 1.8x10-5 Kg/m.s
How long must the chamber be to give theoretical perfect collection efficiency?
Determine the collection efficacy for particles of the same density that are 25 µm in diameter.

48. Making L larger which make the device very long and expansive, by making H smaller which need to subdivide the chamber with horizontal plates which makes the cleanup much more difficult. By lowering Vava requires larger cross sectional area and hence larger and more costly device. OR By increasing g BY substituting some other force for the force of gravity in driving the particles from the gas stream to the collecting surface.

49. Centrifugal separators
The disadvantage of gravity settlers is not effective in small particles. Therefore, it has to find a substitute that is more powerful then the gravity force to drive the particles to the collection surface.

50. Centrifugal separators
If a body moves in a circular path with radius r and velocity Vc along the path then it has angular velocity Ɯ= Vc/r and

51. Centrifugal separators
Gravitational settling velocity less than a hundredth of centrifugal one .

53. Cyclone separator

54. AIR Quality Control Cyclone Collector
Cyclones provide a low cost, low maintenance method of removing larger particulates from a gas stream.
Mechanism
1- The particulate is forced to change direction.
2- The particles continue in the original direction and be separated from the stream.
3- The walls of the cyclone narrow towards the bottom of the unit allowing the particles to be collected in a hopper.
4- The cleaner air leaves the cyclone through the top of the chamber.

55. Cut Diameter
Estimate the cut diameter for a cyclone with inlet width 15 cm, Vc= 18 m/s and number of spiral turns = 5
Particles density 2000 kg/m3 ,
Viscosity 1.8x10-5 Kg/m.s

56. Centrifugal separators
Ex: A particle is travelling in a gas stream with velocity of 18 m/s and radius of 30 cm , what is the ratio of centrifugal force to the gravity force acting on it?

57. Centrifugal separators
Ex: compute the efficiency –diameter relation for a cyclone separator that has Wi = 15 cm , velocity 18 m/s and n=5. for both block and mixed flow assumptions, assuming stokes law

58. Pressure drop
Most of the cyclones, have pressure drops equal to the 8 of velocity heads.

59. Pressure drop
Example : A cyclone has an inlet velocity of 18 m/s and reported pressure loss of 8 velocity heads. What is the pressure loss in pressure units? Assume gas density of 1.20 Kg/m3 .

60. Electrostatic Precipitator ESP
Gravity settlers and centrifugal settlers are devices driving the particles against a solid walls. For particles below 5 micron , the above devices are not effective. ESP is like gravity of centrifugal forces driving the particles to the wall and effective on much smaller particles than the previous devices

61. Electrostatic Precipitator ESP
Which is better relation to diameter ?

62. Electrostatic Precipitator ESP
The basic idea of ESP is to give the particles an electrostatic charges and then put them in electrostatic field that drives them to the collecting wall. Two steps charging and collecting process. Common known electronic air filter.

64. The gasses possesses between the plates and which are electrically grounded (zero voltage). Between the plates are rows of wires held at a voltage of volts. This combination of the charged wires and grounded plates produces both the free electrons to charge the particles and the field to drive them against the plates.

65. For particles larger than 0
For particles larger than 0.15 micron, the dominant charging mechanism is field charging . As the particles become more highly charged , they bend the paths of the electrons away from them. Thus the charge grows with time reaching a steady state value of

66. q is the charge on the particle is dielectric constant of particle is dimensionless, 1 for vacuume, for air and 4 to 8 for typical solid particles. is the permittivity of free space 8.85 x C/V.m. D is the particle diameter and Eo is the local field strength C= x 1019 electrons

67. Ex : a 1 micron diameter particle of a material with dielectric constant of 6 has reached its equilibrium charge in an ESP at a place where the field strength is 300 kV/m. how many electronic charges has it ?

68. The electrostatic force on a particle is Ep is the local electric field strength causing the force Why we do use Ep in this equation and Eo in the previous equation.

69. For both E will apply the average and find the drift velocity Since proportion to particle diameter, one would compute larger values for the larger particles present in the stream. Proportion to the square of E which is approximately equal to the wire voltage divided

70. By the wire to plate distance
By the wire to plate distance. How to achieve unlimited drift velocities? By raising the voltage or lower the wire to plate distance.

71. Ex : a 1 micron diameter particle of a material with dielectric constant of 6 has reached its equilibrium charge in an ESP at a place where the field strength is 300 kV/m. how many electronic charges has it ? Calculate the drift velocityfor the particle.

72. If now consider the section between the row of wires and one plate, its collection area is : The volumetric flow through the section is :

73. Ex : compute the efficiency diameter relation for an ESP that has particles with dielectric constant of 6 and (A/Q) = 0.06 min/m for mixed flow only

74. Ex : ESP has measured efficiency of 90 percent
Ex : ESP has measured efficiency of 90 percent. The target is to achieve 99 percent, by how much must increase the collecting area ?

75. Problem 1 : We whish to design a particle sampling device that will have a greased , horizontal microscope slide 75 mm long over which a contaminated gas stream will flow in a narrow slit 2 mm high. What velocity should we choose if we want 90% of the 1 micrometer diameter particles to fall (and stick ) to the surface of the microscope slide.

76. Problem 2 : Cyclone separator is operating with Dcut= 5 micron
Problem 2 : Cyclone separator is operating with Dcut= 5 micron. It is now necessary to increase the flow rate to the cyclone by 25 percent. Estimate the new cut diameter .

77. Problem 3 : At the equator, what is the ratio of centrfugal force due to the earths rotation to the gravitational force? The radius of the earth is about 6400 km and its rotational velocity at the equator is about 1600 km/hr.

78. Problem 4 : An ESP is treating a particle air stream, collecting 95 percent of the particles. We now double the air flow rate keeping the particle loading constant. What is the new percent recovery. Assume mixed flow.

79. Problem 5 : Exhaust gas from a low efficiency dry ESP at a coal-fired electric power plant. They report average wet ESP penetrations of 5% with A/Q=0.035 min/m. estimate the drift velocity of this wet ESP