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## Fundamental of aerosol: Formation and dynamics

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**Fundamental of aerosol: Formation and dynamics**By GazalaHabib Assistant Professor Department of Civil Engineering, IIT-Delhi**Outline**• What is aerosol? • Life time and transport of aerosol compared to gases. • How do aerosol look like? • Why do we bother about these tiny particles? • Aerosol formation? • Aerosol size and shape • Forces on Aerosol**Residence time and transport of aerosol**Dispersion of Pollutants Introduced into the Atmosphere as Determined by Residence Times (Husar and Patterson, 1980) After formation, the aerosols are mixed and transported by atmospheric motions and are primarilyremoved by cloud and precipitationprocesses.**Size and Shape of Aerosol**• Size range: 0.001 mm (molecular cluster) to 100 mm (small raindrop) Iron oxide particles Granite cutting particle Fly ash particle from coal burning**Aerosol Size Distribution**• - «nucleation: radius isbetween 0.002 and 0.05 mm. Theyresultfrom combustion processes, photo-chemicalreactions, etc. • - « accumulation:radius isbetween 0.05 mm and 0.5 mm. Coagulation processes. • -« fine:particles (nucleation and accumulation) resultfromanthropogenicactivities, • - «coarse: largerthan 1 mm. Frommechanicalprocesseslikeaeolianerosion. 0.01 0.1 1.0 10.0**Visibility Degradation from Aerosols**Glacier National Park, Montana 7.6 µgm-3 12.0 µgm-3 65.3 µgm-3 21.7 µgm-3**What is radiative forcing by aerosols?**DFlTOA DFlSUR**Aerosol and climate change**Top of the Atmosphere (+ve forcing) Knowledge gap: Large uncertainty in quantification of impact of aerosol on climate [IPCC, 2007]. Surface (-Ve forcing) Radiative Forcing (Wm-2) due to aerosol Cloud with aerosol Numerous cloud nuclei Small droplets, Brighter cloud, less prone to rain Drought Annual mean precipitation (1976-2003) minus (1948-1975): Green/blue (red/Yellow) decrease (increase)**Ship Track Formation – the First Evidence of Aerosol**Indirect Effect N ~ 40 cm-3 W ~ 0.30 g m-3 re ~ 11.2 µm N ~ 100 cm-3 W ~ 0.75 g m-3 re ~ 10.5 µm “Borrowed” from Michael King**Aerosol-indirect climate effect Ship tracks off the**Washington coast • Adding CCN makes clouds with more, smaller droplets. • These clouds are whiter, reflect more sunlight net cooling**Formation of aerosol**• Aerosol formation at source: • Primary aerosol formation: Product of incomplete combustion • Elemental carbon • Organic carbon Elemental cabon Organic cabon EC+OC**Formation of aerosol-2**• Aerosolssmallerthan 1 µm are mostlyformed by condensation processessuch as conversion of sulfurdioxide (SO2) gas (releasedfromvolcaniceruptions) to sulfate particles and by formation of soot and smokeduringburningprocesses. • Aerosolparticleslargerthan about 1 mm in size are produced by windblowndust and seasaltfromsea spray and burstingbubbles**Secondary aerosol formation in the Atmospher**Soil dust Sea salt Environmental importance: health (respiration), visibility, climate, cloud formation, heterogeneous reactions, long-range transport of nutrients…**Can you name the laws of motion for a moving object?**• Newton’s law • stokes law**Do aerosol follow Newton’s law?**• Newton’s law • Inertial force dominates the viscous force • What is Reynolds number? • Re>1000 • Large body such as cannonballs (not for particles) • http://csep10.phys.utk.edu/astr161/lect/history/newton3laws.html**Stokes’s Law**• Viscosity dominates over Inertia • http://members.shaw.ca/gp.lagasse/Centrifuge%20Training/basic2.html • Solution of Navier-Stokes equations (differential eqs describing the fluid motion) Navier-Stokes equations are derived from application of Newton’s second law to a fluid element on which the forces include body forces, pressure, and viscous forces.**Stoke’s Law: Assumptions for solving Navier-Stoke equtions**• Resulting equations are very difficult to solve because they are nonlinear partial differential equations. • Therefore, Stoke’s solution involved the assumptions • Inertial force is negligible compared to viscous force: this eliminates the higher order terms in Navier-Stokes equation and yield linear equation that can be solved. • Fluid is incompressible. • There are no wall or other particles nearby • The motion of the particle is constant • The particle is rigid sphere • The fluid velocity at the particle surface is zero. The net force acting on the particle is obtained by integrating the normal and tangential forces over the surface of the particle. Re<1.0 (error in the drag force will be 12% error at Re=1.0 and 5% error at Re=0.3)**Stokes’s law: Assumptions valid or invalid**• Fluid is incompressible • Air around the particle can not be compressed significantly when particle moves through it. Valid • Presence of the wall within 10 diameters of particle will modify the drag coefficient . Aerosols are of small size therefore only a tiny fraction of aerosol will be within 10 particle diameters in any real container or tube. Valid**Stokes Law: Non-rigid particle**• What if it is water droplet (non rigid sphere)? • Settle 0.6% faster than predicted • Reason circulation develop within the droplet caused by resisting force at drop let surface**Aerosol settling by gravity**• Drag force (FD) = Inertial force (FG=mg) pg For water droplet settling in air Not valid for particles less than 1.0 mm size.**Aerosol settling by gravity …contd.**• 10% accurate for particle with standard density having diameter of 1.5-75 mm. • Mechanical mobility of particle (for d>1.0 mm) • Ratio of terminal velocity of particle to the steady force producing that velocity.**Example 1: What are the terminal velocity drag force and**mobility of a 2.5 mm diameter iron-oxide sphere settling in still air? The density of iron oxide is 5200 kg/m3. η=1.81X10-5**What about Particles 0.1 mm to 1.0 mm diameter?**• An important assumption of Stokes’s law is the relative velocity of gas right at the surface of the sphere is zero. The assumption is not met for small particles whose size approaches the mean free path of the gas such particles settle faster than predicted by stokes law because there is a slip at the surface of the particle. At standard conditions, this error become significant for particles less than 1 mm in diameter. • In 1910 Cunningham derived a correction factor for Stokes’s law to account for effect of slip.**What about Particles 0.1 mm to 1.0 mm diameter?**Particle dp<1.0 mm Particle dp>1.0 mm Include slip correction factor or Cunningham correction factor (Cc) = Mean free path (For air at 1 atm 0.066 mm) Slip correction factors for particles 1.0 mm size is 1.15 that means the particle settles 15% faster than predicted by stokes equ**Particles less than 0.1 mm diameter**• For particle less than 0.1 mm • Settling velocity (V): When Re<1.0**Slip correction factor**• Slip correction factor decreases with increase in particle diameter.**Nonspherical Aerosol**• Liquid droplets less than 1 mm and some solid particle are spherical. Most other type of particles are non spherical. • Some have regular geometric shapes, such as cubic (sea salt particles), cylindrical (bacteria and fibers). • Agglomerated particles, crushed material have irregular shape. • Dynamic shape factor (α) is applied to Stoke’s Law to account for effect of shape on particle motion. α is the ratio of actual resistance force of the nonspherical particle to the resistance force of sphere having the same volume and velocity as nonspherical particle. de = equivalent volume diameter**Aerodynamic diameter**Aerodynamic diameter always greater than stokes’s equivalent dia.**When particle is travelling in accelerated field**• This is important for understanding the collection mechanism of aerosols. Such as cascade impactor. • Relaxation time and stopping distance are important • Relaxation time characterizes the time required by the particle to adjust or relax its velocity to anew condition of force. • Relaxation time (τ) = mass X mobility=mB**Stopping Distance**• Maximum distance a particle with an initial velocity V0 will travel in still air in the absence of external force. • S= V0*τ • Velocity of the particle at any time t in accelerating field • V(t)=Vf-(Vf-V0)e-t/τ • For the particle released in still air and accelerating to its terminal velocity Vf is settling velocity. How long will it take a 30 mm glass sphere (p = 2500 kg/m3) to reach a velocity equal to 50% of settling velocity if it is released from rest in still air.**Thermal and radiometric forces**• When the temperature gradient is established in a gas, an aerosol particle in that gas experience a force in the direction of decreasing temperature. The movement of the particle that results from this force is called thermophoresis. • The magnitude of the thermal force depends on gas and particle properties, as well as temperature gradient. • Thermal precipitators are used for aerosol collection. Example in real life: heated metal rod immersed in smoke. The aerosol move away from the rod**Radiometry force**• Photophoresis is a special case of thermophoresis in which the absorption of light by particle creates a temperature gradient in the particle. The gas immediately around the particle takes on the same gradient and establishes the radiometric force.**References**• Hinds, W. C. (1999) Aerosol Technology: Properties, Behavior, and measurement of air born particles. John Willey & Sons Inc. • Friedlander S. K. (2000) Smoke, dust, and haze: fundamentals of aerosol dynamics. Oxford University Press.