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ATMS 455 – Physical Meteorology

ATMS 455 – Physical Meteorology. Today’s lecture objectives: Nucleation of Water Vapor Condensation (W&H 4.2) What besides water vapor do we need to make a cloud? Aren’t all clouds alike?. http://www.artcyclopedia.com/feature-2001-08.html. ATMS 455 – Physical Meteorology.

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ATMS 455 – Physical Meteorology

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  1. ATMS 455 – Physical Meteorology • Today’s lecture objectives: • Nucleation of Water Vapor Condensation (W&H 4.2) • What besides water vapor do we need to make a cloud? Aren’t all clouds alike? http://www.artcyclopedia.com/feature-2001-08.html

  2. ATMS 455 – Physical Meteorology • Today’s lecture topics: • Nucleation of Water Vapor Condensation (W&H 4.2) • Theory • Cloud condensation nuclei

  3. Introduction • Clouds form when air becomes supersaturated wrt liquid water (or ice, in some cases) • Supersaturation most commonly occurs in the atmosphere when air parcels ascend, resulting in expansion and cooling (WH 2.6) • Water vapor condenses onto aerosols forming a cloud of small water droplets Andy Aerosol

  4. Theory • But do we really need (Andy) aerosol to make a cloud droplet? What if we made a cloud via condensation without the aid of aerosols*? Hey! *homogeneous or spontaneous nucleation

  5. Theory • Homogeneous (spontaneous) nucleation • First stage of growth; requires chance collisions of a number of water molecules in the vapor phase to come together, forming small embryonic water droplets large enough to remain intact. Will this happen spontaneously?  Spontaneous implies an irreversible process which implies a total increase in entropy which implies an upper limit on the change in Gibbs Free Energy

  6. Theory • Homogeneous (spontaneous) nucleation (cont.) • Recall: a system (droplet + environment) approaches an equilibrium state by reducing its energy (DE<0) in time

  7. Theory • Subsaturated conditions (e < es) If droplet grows (R increases), then DE>0, this won’t happen spontaneously.

  8. Theory • Subsaturated conditions (e < es) • Formation of droplets is not favored • Random collisions of water molecules do occur, forming very small embryonic droplets (that evaporate) • These droplets never grow large enough to become visible

  9. Theory • Supersaturated conditions (e > es) If droplet grows (R increases), then DE can be positive or negative

  10. Theory • Supersaturated conditions (e > es) - DEinitially increases with increasing R • DE is a maximum where R = r • DE decreases with increasing R beyond R = r

  11. Theory • Supersaturated conditions (e > es) • Embryonic droplets with R < r tend to evaporate • Droplets which grow by chance (collisions) with R > r will continue to grow spontaneously by condensation • They will cause a decrease in the energy (total energy) of the system

  12. Theory • Kelvin’s formula can be used to • calculate the radius r of a droplet which will be in (unstable) equilibrium with air with a given water vapor pressure e • determine the saturation vapor pressure e over a droplet of specified radius r

  13. Theory • Kelvin’s formula can be used to • calculate the radius r of a droplet which will be in (unstable) equilibrium with air with a given water vapor pressure e • determine the saturation vapor pressure e over a droplet of specified radius r • r = 0.01 micrometers requires a RH of 112.5% • r = 1.0 micrometer requires a RH of 100.12%

  14. Theory • Supersaturations that develop in natural clouds due to the adiabatic ascent of air rarely exceed 1% (RH=101%) • Consequently, droplets do not form in natural clouds by the homogeneous nucleation of pure water…

  15. Theory • …droplets do form in natural clouds by the heterogeneous nucleation process • Cloud droplets grow on atmospheric aerosols Yes!

  16. Theory • Droplets can form and grow on aerosol at much lower supersaturations than are required for homogeneous nucleation • Water vapor condenses onto an aerosol 0.3 micrometers in radius, the water film will be in (unstable) equilibrium with air which has a supersaturation of 0.4% Aerosols give a “boost” to the size of a growing cloud droplet.

  17. Theory • Aerosol types • wettable; aerosol that allows water to spread out on it as a horizontal film • soluble; dissolve when water condenses onto them

  18. Theory • Soluble aerosols • solute effect has an important effect on heterogeneous nucleation • Equilibrium saturation vapor pressure over a solution droplet (e.g. sodium chloride or ammonium sulfate) is less than that over a pure water droplet of the same size

  19. Theory • expression may be used to • Calculate the vapor pressure e’ of the air adjacent to a solution droplet of specified radius r • Calculate the relative humidity of the air adjacent to a solution droplet of specified radius r • Calculate the supersaturation of the air adjacent to a solution droplet of specified radius r

  20. Theory • Kohler curve Variation of the RH of the air adjacent to a solution droplet as a function of its radius

  21. Theory • Kohler curve • Below a certain droplet size, the vapor pressure of the air adjacent to a solution droplet is less than that which is in equilibrium with a plane sfc of water at the same temperature • As the droplets increase in size, the solutions become weaker, the Kelvin curvature effect becomes the dominant influence • At large radii, the RH of the air adjacent to the droplets becomes essentially the same as that over pure water droplets

  22. Theory • Focus on curve #2 (solution of 10-19 kg of sodium chloride)

  23. Theory • Curve #2 (solution of 10-19 kg of sodium chloride) Radius of 0.05 mm  RH of 90%  If an initially dry sodium chloride particle of mass 10-19 kg were placed in air with RH equal to 90%, water vapor would condense onto the particle, the salt would dissolve, and a solution droplet of r = 0.05 mm would form.

  24. Theory • Curve #2 (solution of 10-19 kg of sodium chloride) RH of 100.2%  radius of 0.1 mm  If an initially dry sodium chloride particle of mass 10-19 kg were placed in air with RH equal to 100.2%, a solution droplet of r = 0.1 mm would form on the sodium chloride particle

  25. Theory • In both examples the droplets that form are in stable equilibrium with the air since, • if they grew a little more, the vapor pressures adjacent to their surfaces would rise above that of the ambient air and they would evaporate back to their equilibrium size • if they evaporated a little, their vapor pressures would fall below that of the ambient air and they would grow back to the equilibrium size by condensation

  26. Theory • Droplets small enough to be in stable equilibrium with the air are called haze droplets. All droplets in a state represented by points on the left hand side of the maxima in the curves shown in Fig. 4.12 are in the haze state.

  27. Theory • Curve #2 (solution of 10-19 kg of sodium chloride) RH of 100.36% , radius of 0.2 mm (1) Slight evaporation  growth by condensation back to its original size (2) Slight growth  growth by condensation  continued growth  activated droplet (a droplet has passed over the peak in its Kohler curve)

  28. Theory • Curve #2 (solution of 10-19 kg of sodium chloride) RH of 100.4% ambient air RH Growth by condensation, supersaturation of the air adjacent to the droplet would rise. Once droplet reaches peak in Kohler curve, supersaturation of the air adjacent to the droplet would still be below that of the ambient air  droplet continues to grow by condensation.

  29. Theory • Any droplet growing along a curve which has a peak supersaturation lying below the supersaturation of the ambient air can form a cloud droplet (EX1) • Any droplet growing along a Kohler curve which intersects a horizontal line in Fig. 4.12, corresponding to the supersaturation of the air, can only form a haze droplet (2) EX1 EX2

  30. Cloud condensation nuclei • Aerosol which serve as the nuclei upon which water vapor condenses in the atmosphere are called cloud condensation nuclei (CCN). Andy (a.k.a. “CCN”)

  31. Cloud condensation nuclei • CCN types • soluble; the larger the size of an aerosol and the larger its water solubility, the lower will be the supersaturation at which it can serve as a CCN • insoluble; the larger the size of an aerosol and the more readily it is wetted by water, the lower will be the supersaturation at which it can serve as a CCN

  32. Cloud condensation nuclei • For a given environment of 1% supersaturation: • soluble; CCN can be as small as 0.01 mm in radius • insoluble; CCN need to be at least about 0.1 mm in radius

  33. Cloud condensation nuclei • Measuring CCN; thermal diffusion chamber CCN counted using photographs or by measuring the intensity of light scattered by droplets in the chamber

  34. Cloud condensation nuclei • Near the earth’s surface, continental air masses are generally significantly richer in CCN than are marine air masses

  35. Cloud condensation nuclei • Concentrations of CCN over land decline by about a factor of five between the sfc and 5 km • Concentrations of CCN over the ocean remain fairly constant with height

  36. Cloud condensation nuclei • CCN source region is over land • Soil and dust particles are not dominant • Forest fires are sources of CCN • Sea-salt particles are not a primary source of CCN • Gas-to-particle conversion mechanisms might be important sources of CCN • Many CCN consist of sulfates

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