EN3004 Air Pollution Control Engineering Week 9 PDF

EN3004 Air Pollution Control Engineering Week 9 Properties of particulate matters & Particle collection mechanisms Tuti Lim School of Civil and Environmental Engineering 12 Nanyang Technologic...

EN3004 Air Pollution Control Engineering Week 9 Properties of particulate matters & Particle collection mechanisms Tuti Lim School of Civil and Environmental Engineering 12 Nanyang Technological University 1  Outline (I)Properties of Particulate Matters ⚫ Formation and Characteristics of Particulate Matters ⚫ Particulate size and size distribution (II) Forces acting on particles & Stokes Law (III) Particle Collection Mechanisms ⚫ Impaction ⚫ Interception ⚫ Diffusion 2  Properties of Particulate Matters PM – sums of all solid and liquid particles suspended in air, many of which are hazardous. 3  Common PM Pollutants Solid particles larger than colloidal size capable of Dust temporary suspension in air Finely divided particles of ash entrained in flue gas. Fly ash Particles may contain unburned fuel Particles formed by condensation, sublimation, or Fume chemical reaction, predominantly smaller than 1g (tobacco smoke) Dispersion of small liquid droplets of sufficient size to fall Mist from the air Smoke Small gasborne particles resulting from combustion Particle Discrete mass of solid or liquid matter Fog Visible aerosol Soot An agglomeration of carbon particles 4 A3 How Particulate Pollutants are Formed? 1: Condensation processes atmospheric reactions combustion 2. Accumulation processes 3 combustion 2 coagulation 1 condensation on existing particles Rainout atmospheric reactions and Sedimentation Washout 3. Mechanical processes wind blown dust Emissions Particle in air are either: Sea spray 1. Directly emitted (primary particles) – 3 peak. Volcanoes rd 2. Indirectly formed from gaseous precursors in Plant Emitted Particles the air (secondary particles) – 1st & 2nd peaks. 5 A3 How Particulate Pollutants are Formed? Finest particles (0.005-0.1μ) enters the atmosphere by condensation of hot vapors. Overtime they grow by agglomeration onto each other either in gas phase or inside cloud or fog droplets. Midsize particles (0.1-1μ) formed by agglomeration and/or chemical conversion of gases and vapor in the air. 3 They can be removed by 2 rainout (drops in clouds) or 1 washout (rainfall). Rainout Agglomeration process is and Sedimentation slower than rainout & Washout washout. Coarse particles (2-100μ) are mechanically generated and can be removed by capturing device such as gravity settler. 6 Coagulation: Particles collide and stick together. Condensation: Gases condense onto a small solid particle to form a liquid droplet. Cloud/Fog Processes: Gases dissolve in a water droplet and chemically react. A particle exists when the water evaporates. Sulfate Chemical Reaction: Gases react to form particles. Source: Timothy S. Dye, et al. 7 A3 video video In the area close to the exhaust the aerosol shows high concentration and narrow size distribution (primary particles). As the air carries the aerosol away from the exhaust, the particles collide with each other forming larger agglomerates (decreasing the number of smaller particles.) This process also causes broadening of the size distribution. 8  Characteristics of Particulate Matters The efficiency of the particle collection mechanisms strongly depends on particle size. The particle size distribution of flue gas determines the operating conditions necessary to collect the particles and control device’s collection efficiency (control concern). Particle size distribution are important in determining the behavior of particles in the respiratory tract (health concern). s9 s9 dp > 10 m 1 < dp 10 m PM2.5 Coarse 2.5 m < dp < =10 m Particles less than 0.1 μm Fine 0.1 m < dp 50 0 0 1 0 2 0 3 0 4 0 5 0 Total 1000 d p i(m) Q: Which size range has the most particles? 18 A1 Standardized frequency/dp vs Particle Size dpi fi(dpi) Pro(babifractliitoyn/Densidtpiy)Function Size Range Fraction/size 0. 08 (m) (1/m) 0-4 0.026 0. 06 4-6 0.08 6-8 0.0805 8-9 0.075 0. 04 9-10 0.067 10-14 0.465 0.0465 14-16 0.0305 0. 02 16-20 0.0197 20-35 0.0034 0. 00 35-50 0.0001 0 1 0 2 0 3 0 4 0 5 0 > 50 0 0 n d i(m) fi = i p N 19 A1 Continuous Particle Size Distribution If the size range is very small, the discrete PSD will approach continuous PSD. f df q( d p ) = i = d pi dd p  →0 Lognormal distribution for size distribution of d/dD particulate matters has been widely used for aerosol science. dp : Cumulative Fraction of particles(0 ~ D) d 1  (ln D − ln Dmean, g ) 2  D: Diameter of particles = exp −   Dmean,g Geometric Mean Dia. d ln D ln  g 2  2(ln  g ) 2  g Standard deviation 20 A7 PM Size Distribution: Representative Size ⚫ MEAN (arithmetic average): Dmean 1 Dmean =  D = n1D1 + n2 D2 + n3 D3 +... = n D i i N n1 + n2 + n3 +... n i ⚫ GEOMETRIC MEAN (Logarithmic mean): Dmean, g Dmean, g = ( D1 1 D2 2 D3 3......)1 / N n n n Expressed in terms of ln (Dmean, g): ln( Dmean, g )=  n ln D i i Dmean, g = exp(  n ln D ) i i N N GEOMETRIC MEAN diameter is also called MEDIAN diameter, for which 50% of the total are smaller and 50% are larger; the diameter corresponds to a cumulative fraction of 50%. Expressed as D50%. 21  PM Size Distribution: Standard Deviations ⚫ Standard deviation:  (A measure used to quantify the amount of variation or the dispersion of a set of numbers).   ( D − Dmean )   ni ( Di − Dmean ) 1/ 2 1/ 2 2  2  =  =   N −1   N −1  ⚫ Geometric (logarithmic) standard deviation: g   ( D − Dmean )   ( D − Dmean ) 1/ 2  1/ 2 2 2  ln  g =    g = exp    N −1   N −1  = exp() 22  Features of Lognormal Distribution d 1  (ln D − ln Dmean, g ) 2  = exp −   log( d8 4.1 ) = log( d5 0) + log  g d ln D ln  g 2  2(ln  g ) 2  log( d8 4.1 ) − log( d5 0) = log  g d50 d d15.9 d84.1 log  g = log ( 8 4.1 ) d5 0 and log( d1 5.9 ) = log( d5 0) − log  g σg d5 0 log  g = log( d5 0) − log( d1 5.9 ) = log ( ) d1 5.9 Where d84.1 = diameter, particulate constituting 84.1% of the total mass of particles are smaller than this size d84.1 d50 d50 = geometric mean diameter g = = d15.9 = diameter, particles constituting 15.9 % of the d 50 d15.9 total mass of particles are smaller than this size σg = geometric standard deviation Symmetry 23 A1 Data presented in a log-normal figure d84.1 d50 g = = d 50 d15.9 the value of d50, d84.1 and d15.9 thecan be read as 7.8, 13.5 and 4.6 microns. relationship g = 7.8/4.6 = 1.70 or g = 13.5/7.8 = 1.73 between cumulative fraction and diameter of particles is Average linear g = 1.72; d50 = 7.8 m 24  Example A waste stream contains particles of three sizes: large, medium and small. These are present in equal quantities by weight in the gas stream. We pass this gas stream through a collector that is 99% efficient on large particles, 75 % efficient on the medium particles, and 30% efficient on small particles. What is the overall weigh percent efficiency of this collector? Solution: Incoming Mass Collection Mass fraction x efficiency amount fraction, mi Efficiency, ηi mi ηi Large 0.334 kg 0.334 0.99 0.334 x 0.99= 0.33 Medium 0.333 kg 0.333 0.75 0.333 x 0.75 =0.25 Small 0.333 kg 0.334 0.30 0.333 x 0.30 = 0.10 Total 1 kg 1 0.68 Overall efficiency =0.680 25  Example ALTERNATIVELY A waste stream contains particles of three sizes: large, medium and small. These are present in equal quantities by weight in the gas stream. We pass this gas stream through a collector that is 99% efficient on large particles, 75 % efficient on the medium particles, and 30% efficient on small particles. What is the overall weigh percent efficiency of this collector? Solution: Incoming amount Penetration Outgoing amount Large 0.334 kg 1-99%=0.01 0.334 x 0.01= 0.00334 kg Medium 0.333 kg 1-0.75%=0.25 0.333 x 0.25 = 0.0833 kg Small 0.333 kg 1-30%=0.70 0.333 x 0.70 = 0.2331 kg Total 1 kg 0.3197 kg Total penetration: p=0.3197/1 =0.320 Overall efficiency =1- p=0.680 26  Efficiency and Penetration Incoming amount Outgoing amount Collector Incoming amount – Outgoing amount Efficiency,  = Incoming amount Outgoing amount Penetration, p = Incoming amount p = 1-  the fraction not collected 27 Total penetration: p = p1x m1 + p2 x m2 + p3 x m3 +…… = (1- 1) m1 + (1- 2) m2 + (1- 3) m3 +…… = (m1 + m2 + m3 +……) -(1 m1 + 2 m2 + 3 m3 + …..) = 1 - (1 m1 + 2 m2 + 3 m3 + …..) Overall efficiency  = 1- p = 1 m1 + 2 m2 + 3 m3 + …..  =  j m j S29 S29 Where: j = efficiency of collection for the j size range mj = mass percent of particles on the j size range When the particulate size distribution is known, and the efficiency of the device is known as a function of particle size, the overall collection efficiency can be predicted using above equation. 28 Total penetration: p = p1x m1 + p2 x m2 + p3 x m3 = (1- 1) m1 + (1- 2) m2 + (1- 3) m3 = m 1 - 1 m 1 + m 2 - 2 m 2 + m 3 - 3 m 3 = (m1 + m2 + m3) -(1 m1 + 2 m2 + 3 m3) = 1 - (1 m1 + 2 m2 + 3 m3 + …..) Overall efficiency  = 1- p = 1 m 1 +  2 m 2 + 3 m 3 m = mass fraction 29 (II) Forces Acting on Particles & Stokes Law 30  Forces Acting on Particles in a Fluid Assuming a single particle, no the electrostatic buoyancy and Van der Waals.  Forces between the particle and other particles p a rt  air can be ignored. ρpart＞ρair gravity Initially, v = 0 31 A6 Drag force buoyancy Particle falls down, getting an acceleration a  p a rt  v: 0 v air Drag force takes place. (the forces that the surrounding air or gas exerts on a particle) gravity Buoyancy force takes place. (upward forces exerted by fluid to 1. Drag force counterbalance effect of gravity acting on the fluid) 2. Buoyant force 3. Gravity force 32 A6 Drag force buoyancy If Reynolds number is small, the drag force is expressed as   Stokes’ Law FDd = 3 d pV Stokes’ Law (1) a p a rt air gravity Writing Newton's law for the particle, we obtain (2) The drag force increases as the velocity increases until it equals the gravity minus the buoyancy (a =).  Fd = ( )d p3 g (  particle −  fluid ) (3) 6 (  particle −  fluid ) Vt = gd 2 18 p (4) At this time, the velocity is terminal settling velocity, the sum of the forces acting is zero, so the particle continues to move at a constant velocity. 33 A2 Terminal Settling Velocity 0 If  particle >>  fluid (  particle −  fluid ) Vt = gd 2 18 p Terminal settling velocity gD2 2  gd pparticle Vt = part 18 Units: g = 9.81 m/s2 dp - m  - kg/m3  - kg/m.s = 1.8  10-5 kg/m.s at 20C 34 A4 Why are We Interested in Settling Velocity? gd22pparticle gD Vt = part Vt ∝ d D22 18 p 1 μm Vt = 0.006 cm/s 10 μm Vt = 0.6 cm/s Vt 1 μm = 10-6 m 1000 μm Vt = 6000 cm/s u = Vertical wind velocity For smaller particles, u > Vt Suspending in the atmosphere for a long time. For larger particles, u < Vt Particles will settle to the ground, not causing air pollution. 35  Assumptions for Stokes’ Law Stokes’ Law was derived with assumptions: (1) Fluid is continuous (particles are larger than spaces between individual gas molecules/atom, fluid interacts with particle as if it is continuous medium) (2) Flow is laminar (3) Newton’s law of viscosity holds (4) The terms involving velocities squared are negligible. 33 33 36 A2 Vt actual ＜Vt Stokes λ – mean free path Valid for range of Ave travel dist of gas conditions which 0.3≤Re≤1000 molecules between assumption holds. collision Stokes Law 1 gD  part 2  4 D  part g  2 Vt = Vt =  3  18  Cd  fluid    Cd = (1 + 0.14Re0.7 ) 24 Re DV  fluid 1.728 Re = ~0.07μm  C=1+Aλ/D 1. Fluid is continuous 2. Laminar Flow Fd gD 2  part C 3. Newton’s law of Cd = Vt = = VStokes C  V 2  ) D  fluid  18 viscosity holds ( 4 2  4. Negligible v2  2  S38  Vt S38 S41 S41 Vt actual Stokes S39 S39 37 A1 Particles too Large for Stokes’ Law Stokes’ law becomes inapplicable as the flow is turbulent DV  fluid Reynolds number Re = (1)  Laminar flow Re  0.3 stokes’ law works satisfactorily 0.3  Re  1000 flow becomes turbulent. Re – a dimensionless ratio of the inertial forces acting on the mass of the fluid to the viscous forces acting on the same mass of the fluid in the same flow. 38 A7 Fd Drag Coefficient Cd = (1) V  2 ( 4 ) D 2  fluid    2  At the terminal   3 Fd =   D  part g (2) 6 Combining (1), (2), (3) settling velocity Re Number Drag Coefficient Settling velocity 24 gD 2  part Vt = (3) Re