1. Lecture 2: Composition & Structure of the Atmosphere (Ch1) Composition (continued) Vertical structure of the atmosphere Hurricane Florence, a problem for Bermuda on Monday, was expected to bring strong winds and heavy rain to eastern Newfoundland on Wednesday. (NOAA)… by that time, Florence will probably be gearing down from a hurricane to a "very intense post- tropical storm," Environment Canada's Canadian Hurricane Centre said in a statement issued early Monday.

2. O 3 (ozone) and the ozone hole radiatively/climatologically active (accounts for peak T in mid-atmosphere) inactive relative to weather essential relative to life… absorbs ultraviolet radiation peak concentration of up to about 15 ppm in mid-stratosphere depletion in spring over poles (esp. Antarctica) reactions on surface of polar stratospheric ice clouds over long term, size of ozone hole governed by emissions of CFCs, lifetime order 100 years year-to-year variability determined mostly by temperature variations (NASA) Cl + O 3  O 2 + ClO ClO + O  O 2 + Cl

3. Aerosols size 0.1  m  100  m or larger (smallest formed from sulphate gases) influence visibility (VV see Appendix C) increase shortwave reflectivity trap outgoing longwave radiation form cloud condensation nuclei UA farm Ellerslie 28 May 2001 * An uncertain feedback in climatic modelling: DMS (dimethyl sulphide) gas released by decay of ocean biota generates aerosol with radiative impact as well as acting as cloud condensation nucleii (CCN) 1 o C reduction in N. hemisphere sfc temp a year after Pinatubo eruption 1991

4. Vertical Structure Fig 1-8 Fig 4-3 Density (  ) and pressure (p) both decrease with increasing elevation

5. Air column, base area A p=0 Ground- or sea-level p=p 0 Top of atmosphere p 0 = weight of air divided by area A = M g / A [Pascals] Relates to Fig. 4-2

6. Vertical Structure We shall neglect these layers Fig. 1-9

7. Ideal Gas Law (equation of state) (p96, 4 th edition) p, pressure [Pa] , density [kg m -3 ] R = 287 [J kg -1 K -1 ], specific gas const. for dry air T, temperature [K]

8. Question: these paragliders are flying at a height of 1000 m above sea-level. A pilot’s instrument reports so the air density at flight level is?

9. The Hydrostatic Law Gives the change in pressure (  p) associated with an increase (  z) in height  = air density [ kg m -3 ] (approximately 1, near ground) g = grav. accel’n = 9.81 [ m s -2 ] (approximately 10) Thus: Thus: near ground, pressure increases by an amount  p = - 10 Pa for each 1 m increase in height 100 Pa (= 1 mb)per 10m 100 mbper 1 km (p104, 4 th edition)

10. Question: if those paragliders descend 100 m, estimate the pressure at their new flight level: p 1, T 1 (known) p 2,T 2  To find: p 2

11. Question: if those paragliders descend 100 m, estimate the pressure at their new flight level: p 1, T 1 (known) p 2,T 2  To find: p 2 Need to use with  z = - 100 m. But we need the density  …

12. Approximate the density as  (note the unit conversions; we work in MKS units) (implies) A negative step in height z gives us a positive step in pressure p