Transcript Slide 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 posttropical storm," Environment Canada's Canadian Hurricane Centre
said in a statement issued early Monday.
O3 (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
Cl + O3  O2 + ClO
ClO + O  O2 + Cl
• 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)
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)
1oC reduction in N. hemisphere sfc temp a year after Pinatubo eruption 1991
Vertical Structure
Fig 4-3
Fig 1-8
Density () and pressure (p)
both decrease with
increasing elevation
p=0
Top of atmosphere
Air column,
base area A
Relates to Fig. 4-2
p=p0
Ground- or sea-level
p0
= weight of air divided by area A = M g / A [Pascals]
Vertical Structure
Fig. 1-9
We shall neglect these layers
Ideal Gas Law (equation of state)
p R T
p
(p96, 4th edition)
, pressure [Pa]
 , density [kg m-3]
R = 287 [J kg-1 K-1] , specific gas const. for dry air
T , temperature [K]
Question: these paragliders are
flying at a height of 1000 m above
sea-level. A pilot’s instrument
reports
p  900 [mb],
T  25[ C]
o
so the air density at flight level is?
The Hydrostatic Law
(p104, 4th edition)
Gives the change in pressure (p) associated with an increase (z) in height
p
 g
z
 Pa
1 
or
Pa
m
 m

= air density
[ kg m-3 ]
g = grav. accel’n = 9.81 [ m s-2]
(approximately 1, near ground)
(approximately 10)
Thus: near ground, pressure increases by an amount p = - 10 Pa
for each 1 m increase in height
100 Pa (= 1 mb)
100 mb
per 10m
per 1 km
Question: if those paragliders
descend 100 m, estimate the
pressure at their new flight level:
p1, T1 (known)
p1  900 [mb],
T1  25[o C]
To find: p2

p2,T2
Question: if those paragliders
descend 100 m, estimate the
pressure at their new flight level:

p1, T1 (known)
p2,T2
p1  900 [mb],
T1  25[o C]
To find: p2
Need to use
p
 g
z
with z = - 100 m. But we need the density  …
Approximate the density as
p1
900x 102
  1 


RT1 287 273 25
(note the unit conversions; we work in MKS units)
p
 g
z

(implies)
p   z 1 g  100 1 10
A negative step in height z gives us a positive step in pressure p