The atmospheric compartment

1. The atmospheric compartment • How much does it weigh? • Temperature and pressure • Circulation and mixing • Where did Oxygen come from • Particle emissions • Emissions of other pollutants

2. The total atmosphere weights5.13x1018 kg Assume that an average car weighs 1000 kg or one metric tonnehow many cars are equal to the weight of the atmosphere?if you could cover the earth in cars how high would the stack be? Assume the average car has a “foot print” of 4x2 meters and is 2 meters high?

4. Two important features the atmosphericCompartment aretemperature and pressure

5. The atmosphere is usually divided into the following: • Troposphere 0-10 km • tropopause ~10km • Stratosphere 10-50km • stratopause ~50 km • mesosphere 50-80kn • Thermosphere 80 + km

6. These divisions come about because of temperature differences as one increases in altitude:

7. The troposphere contains about 80% of the atmospheric mass. • Air cools with altitude in the troposphere. The top; 10-15 km is at ~-60oC; which means very little water vapor. • In the stratosphere, temp. increases with height because O3 absorbs uv radiation. • Thermal mixing of air (heat) is responsible for global circulation in the lower atmosphere.

8. The atmosphere is held to earth’s surface • by the gravitational attraction of the earth • At a given altitude the downward force (F) is related to the mass (M) of the atmosphere above that point. F= M (g); where g is the gravitational acceleration constant

9. The pressure or force per unit area • decreases with increasing altitude • The decline in pressure (P) with altitude is approximately = to log P= - 0.06 (z); where z is the altitude in km and P is bars

10. How thin is the air at the top of Mt. Everest? • Mt. Everest is 8882 meters high or 8.88 km high • log P = -0.06 x 8.88 • P = 10-0.06x 8.88 = 0. 293 bars • Assume there are 1.01bars/atm. • This means there is < 1/3 of the air

11. Where does log p = - 0.06 (z) come from; • Force = mass x accelerationacceleration =g • The mass of air over a surface,A,equals height x A x mass/volume; the mass / volume = density, r; • So Force = -z x A xrxg • The change in force at any altitude • dF = -dz x A xrxg; F/A = pressure, p So the change in pressure with height isdp= -dz xrxg

12. dp = -dz xrxg • What is the ideal gas law • pV = nRT • Show that r = Mw xp/RT • Substituting for r in dp = -dz xrxg • dp = -dz xMw xp/RTxg • Sodp/p = -dz xMw /RTxg • Integrating • So p = po exp {-Mw g z/RT}

13. p = po exp {-Mw g z/RT} • If we set H = (Mw g/RT)-1 it has the units of length • and we get a simple expressionp= po exp {-z/H} • Solving for H at 290K; R = 8.3 joules/(K mole) one joule = 1kg meter2/sec2average Mw of air = 28.9 g/moleg = 9.8 meter/sec2 • H = 8.5 km; people actually find that 7km works best; when 7 km is used we end up with • log p = - 0.06 (z);

14. Is it possible in the troposphere to calculate the rate that temperature of the air decreases with altitude?

15. To do this we need to start with simple thermodynamics • The first law of thermo says that the change internal energy of a system is the sum of its changes in heat content and work that is done. • dU = dq - dw • A change in work can only occur if a force moves through a distance; dw = d (fxz) = d (pV) for work there must be movement • Hence dw = Vdpand dU = dq - Vdp • Another form of energy is call enthalpy (H) which is the sum of the internal energy and pV from pV=nRT • so H = U + PV or dH = dq –Vdp+ Vdp +pdV = Vdp(dq is assumed to be zero for a process that does not have a heat loss) • The change in the heat of a mass, per change in a degree centigrade, is called its specific heat Cp and Cp = dH/dT

16. we said the enthalpy dH= Vdp • specific heat capacity; Cp = dH/dT • So Cp dT= dH = Vdp • Before we said that the change in pressure with height, z was dp= -dz x r x g • So substituting for dp we get • Cp dT= - V xdz x r x g • So the change in temp with height is dT/dz = - V x r x g/ Cp • Density r is mass/V; so for an air mass of one gram, r = 1/V • This puts Cp in units of energy gram-1 deg-1; for dry air this is 0.24 cal gram-1 o K-1 and we will call it cp • So - dT/dz = g/ cp where g = 9.8 m sec-2 • 1 cal = 4.1 joules so cp= ~1 joule gram-1 o K-1 • One joule = 1 kg m2sec-2 • So - dT/dz = g/ cp = 0.0098 oK/meter or 9.8 oK/kilometer

17. The fact that - dT/dz = g/ cp = 9.8 oK/kilometer is constant is consistent with observations • And this is called the dry adiabatic lapse rate so that - dT/dz = d • When - dT/dz > d the atmosphere will be unstable and air will move (convection) to re-establish a stability

18. The quantity dis called the dry adiabatic lapse rate • Air that contains water is not as heavy and has a smaller lapse rate and this will vary with the amount of water • If the air is saturated with water the lapse rate is often called s • Near the surface sis ~ -4 oK/km and at 6 km and –5oK/km it is ~-6K/km at 7km high

19. Sun-down earth cools more cooling at surface at night midday   altitude altitude altitude } Inversion layer temp temp temp At midday, there is generally a reasonably well-mixed layer lying above the surface layer into which the direct emissions are injected. As the sun goes down, radiative cooling results in the formation of a stable nocturnal boundary layer, corresponding to a radiation inversion.

21. What happens to the material above the inversion layer?? more cooling at surface at night } residual layer altitude } Inversion layer temp These materials are in a residual layer that contains the species that were well-mixed in the boundary layer during the daytime. These species are trapped above and do not mix rapidly during the night with either the inversion boundary layer below or the free troposphere above.

22. more cooling at surface at night Heating at surface during the next day altitude altitude } Inversion layer } Inversion layer temp temp When the sun comes up the next day it heats the earth an the air close to the earth. During the next day heating of the earth's surface results in mixing of the contents of the nocturnal boundary layer and the residual layer above it

24. Mixing height in the morning • We will start with the balloon temperature curve that is taken at the airport each morning. • In the morning the temperature usually increases with height for a few hundred meters and then starts to decrease with height (see the green curve) according to the temperature sensor on the balloon • The the break in the curve usually defines the inversion height in the early morning

25. Balloon temperature } Inversion height Mixing height in the morning height in kilometers Temp in oC

26. Mixing height in the morning • There are another set of lines called the dry adiabatic lines, which are thermodynamically calculated, and represent the ideal decrease in temperature with height for dry air starting from the ground. • In the morning, the mixing height is estimated by taking the lowest temperature just before sunrise and adding 5oC to it, and then moving up the dry adiabatic line at that temperature until it intersects the balloon temperature line or the green curve. • Let’s say the lowest temperature just before sunrise was 20oC. We would add 5oC to it and get 25oC. We then move up the 25oC dry adiabatic line. We then go straight across to the right, to the height in kilometers and get a morning mixing height of ~350 meters (0.35 km). This is illustrated in the next slide. It is animated so you can see it more easily

27. Balloon temperature 1.5 1.1 height in kilometers Dry adiabatic lines 0.4 0.3 0.2 0.1 0.0 20 25 30 35 Temp in oC Mixing height in the morning

28. Mixing height in the afternoon • To get the mixing height in the afternoon, you just take the highest temperature between 12:00pm and 15:00 pm • Do not add anything to it, but as before run up the dry adiabatic curve and intersect the morning balloon temperature curve. • Let say the highest afternoon temperature is 35oC, we would estimate an afternoon mixing height of ~1.67 km

29. Balloon temperature 1.5 1.1 height in kilometers Dry adiabatic lines 0.4 0.3 0.2 0.1 0.0 20 25 30 35 Temp in oC Afternoon Mixing height

30. How does air circulate • At the equator air is heated and rises and water is evaporated. • As the air rises it cools producing large amounts of precipitation in equatorial regions. • Having lost its moisture the air mass moves north and south. • It then sinks and compresses (~30oN and S latitude) causing deserts

31. Circulation currents 30oN Hadleycell equator

32. A similar system at the poles occurs where cold air sinks and flows south (in the northern hemisphere. • These sinking columns of air create a circulation system called Hadley cells

33. Circulation currents Hadleycell indirectcells 30oN Hadleycell equator

34. In between the Hadley cells, north and south of the equator, and at the poles, counter current or indirect cells are set up. • These drive circulation between 40-60o latitude, producing westerly winds and storms

35. Circulation currents Hadleycell indirectcell 30oN Hadleycell equator

36. The air in each hemisphere mixes with a time constant , t, of a few months. • The air between the north and south hemispheres completely mix on the order of one year. • Air mixes into the stratosphere from rising Hadley cells in the tropics, storms and eddy diffusion. • exchange between the troposphere and the stratosphere can be thought of in terms of mean residence times (MRT)

37. The mean residence time (MRT) can be expressed as: MRT = mass / flux where flux is mass/time • If 75% of the mass/year in the stratosphere comes from the troposphere • 1 MRT = ----------------- = 1.3 years • 0.75/year

38. Mt. Pinatubo in the Philippines erupted in June 1991, and added a huge amount of SO2 and particulate matter the stratosphere. After one year how much SO2was left? • For a 1st order process C= Coe-1 year/ MRT • C/Co= e -1 year/ MRT = e -1/1.3= 0.47 or ~ 50% • in 4 years, C/Co= e -4 years/1.3 years = ~5%

39. What happened to global temperatures after the Pinatubo eruption • A lot of SO2 was injected into the atmosphere • What we will learn later that SO2 forms fine sulfate particles that reflect light back into the atmosphere and this cools the upper troposphere

41. Atmospheric Composition • Three gases, O2, N2, and argon make up 99% of the atmosphere mass of 5.14x1021 g • These gases are relatively un-reactive and their mean residence times are much longer than the rate of atmospheric mixing. Hence the conc. of N2, O2, and the Nobel gases (He, Ne, Ar, Kr, and Xe) are relatively uniform.

42. Atmospheric Composition • N2 78.084% 3.87x1021 grams • O2 20.946 1.19x1021 • Ar 0.934 6.59x1021 • CO2 0.036 2.80x1018 • Ne 18.2 ppm 6.49x1016 • H2 510 ppb 1.82x1014 • CFC 11 280 ppt 6.79x1012 • MeBr 11 ppt 1.84x1011

43. Where does oxygencome from in our atmosphere? • ~3.8 billion years ago the earliest bacteria were able to take acetic acid and metabolize it to CO2 and water. • CH3COOH ->CO2+ H2O • A later form of bacteria could obtain energy from the reduction of H2S to S

44. CO2 +2H2S->CH2 O+ 2S + H2O • As supplies of H2S were consumed in the oceans other energy generating metabolic processes became more competitive • one was photosynthesis • H2O +CO2 -> CH2O + O2

45. A good summary of the “Rise of Life on Earth” • is given in National Geographic, vol 193, p 54, March 1998 • Experiments in the early 1950’s (Dr. Stanley Miller, U. Cal. San Diego) showed that is is possible to generate amino acidsin an atmosphere of CH4, NH3 and H2 over a pool of water with an electrical discharge

46. This gave rise to the theory that “life” could have originated in warm tidal pools of the oceans, • since amino acids are the building blocks of “life”. • It is now thought that the atmosphere 4-3.5 billion years ago consisted of CO2 and N2 • sparking CO2 and N2does not generate large amounts of organics

47. A more recent theory is that “life” could have originated in deep hot pools of water heated by volcanic rock. • Some of the most primitive live forms, called thermophil bacteria, can live at 190oC and consume iron and sulfur (3.8 billion years ago). • Another theory involves ice. Trapped H2C=O, NH3, HCN and water--->glycine (amino acid) followed by meteorite impact

48. Photosynthetic cyanobacteria (blue green algae) may go back 3.46 billion years. They were recently found in a rock from NW Australia. • It is felt these types of bacteria invaded other cells and their chloroplasts“stayed” there. • Some bacteria were also invaded by mitochondria which burn sugar

49. These photosynthetic cells gave off huge amounts of oxygen. • Oxygen did not really start to build up in the earth's atmosphere until ~2 billion years ago. Why?? • The oceans contained much dissolved iron. It had to be oxidized before O2 could persist in the atmosphere.

50. Aerosols • The atmosphere contains particles which are known as aerosols. These come from a number of different sources • Soil minerals are dispersed by wind erosion from arid (dry) and semi-arid regions • Particles with a diameter of 1.0 mm (the smallest we can see is 20 mm) are held aloft by turbulent motion and can be transported long distances