Transcript Next 4 weeks: Atmospheric temperature profiles Stability

Next 4 weeks:
Atmospheric temperature profiles
Stability
[1-week on Fronts from Hugh Pumphrey]
Thunderstorms
Air Pollution
David Stevenson
Crew Building Room 314
[email protected]
L13 Physics of Dry Air
• Vertical pressure gradient through the
atmosphere:
– The Hydrostatic Equation (revision)
• Measuring temperature profiles: radiosondes
• 1st Law of Thermodynamics: conservation of
energy
• ‘Air parcels’
• The ‘Dry Adiabatic Lapse Rate’
Why are vertical profiles important?
• At what height do clouds form?
• Will air rise or sink?
Atmospheric stability
=> Rain showers/
thunderstorms
caused by moist
air rising and cooling
Inversions – e.g. fog
Pollution dispersion –
e.g. Buncefield fire
Vertical pressure gradient
• We know pressure (p) decreases as you
go up through the atmosphere.
• Q: So why doesn’t air flow from high p at
surface to low p at altitude? (i.e. why
hasn’t Earth lost its atmosphere to space?)
• A: Gravity attracts it towards the centre of
the Earth.
• The balance of gravity and vertical
pressure gradient is ‘hydrostatic balance’
Hydrostatic Equation (see Lecture 8)
Net upward force
on slab, due to the
pressure gradient
= -p
Downward force on air
in shaded slab, due to
pressure of air above
must balance
downward
force due
to weight
of slab
= gz
Upward force on air in
shaded slab due to
pressure of air below
 p  gz
In limit as
z → 0,
p
  g
z
First law of thermodynamics
Conservation of energy for a parcel of air:
dq  du  dw
dq = heat added to
air parcel from its
surroundings
du = Change in
internal energy
of air parcel
( cpdT )
dw = Work done
by air parcel
 dp
Cp is the specific heat capacity at constant pressure
α is the specific volume (1/density)
dq  cpdT  dp
Pumping up/letting down a tyre
• Pump up tyre: air compresses, work is done on
the air in the tyre (dw is –ve; the air in the tyre
doesn’t do work). If we assume dq=0, then dT is
+ve, the air heats up (valve gets hot).
• Let down tyre: air expands, the air in the tyre
does work on its surroundings (dw is +ve). If we
assume dq=0, then dT is –ve, the air cools down
(valve gets cold).
dq  cpdT  dp
Radiosondes
Temperature (and humidity,
pressure) sensor, attached to
weather balloon, with radio
transmitter to send data back
to earth.
How do we measure temperature
and moisture in the atmosphere?
Radiosonde
• pressure, Temperature and
moisture are all measured by
sensors and the signals are
transmitted to a base station by
radio
• Rises to between 20-30 km
• then balloon bursts and radiosonde
returns to surface by parachute
Ikarus project:
http://www.youtube.com/watch?v=MCBBRRp9DOQ&NR=1
Wind speed and direction
are not directly measured
but inferred
(radar echo,
onboard radio receiver
or GPS-based systems).
All measurements in the profile
are attributed to the nominal hour
of the ascent. This is the hour at
which the sonde reaches 100 mb.
It takes approximately an hour for
the balloon to rise to this level and
thus the sondes are released one
hour before the synoptic hours.
Automatic balloon releases
Radiosonde data
Radiosonde data is reported
up to four times per day at the
synoptic hours of 00, 06, 12
and 18 GMT. The number of
ascents varies widely between
countries and stations.
You can get worldwide
radiosonde data from:
http://weather.uwyo.edu/upperair/sounding.html
Temperature of an ascending air parcel
• Start with the 1st Law of thermodynamics:
dq  cpdT  dp
• Assume the ascent is adiabatic, i.e. dq=0
• Use the hydrostatic equation: dp  
• Gives:
gdz
0  cpdT   ( gdz)
0  cpdT  gdz
or:
remember:
dT  g

dz c p

1

Dry adiabatic lapse rate
Acceleration due to gravity, g = 9.81 m s-2
Specific heat capacity dry air (at constant P), Cp = 1004 J K-1 kg-1
So:
dT  g  9.81
1
1


 0.0098Km  9.8Kkm
dz c p
1004
Check units: remember a Joule, J, can be expressed in
fundamental SI units (e.g., kinetic energy = ½ m v2):
1 J = 1 kg (m s-1)2 = 1 kg m2 s-2
So units of Cp, J K-1 kg-1 = kg m2 s-2 K-1 kg-1 = m2 s-2 K-1
So the temperature gradient has units:
dT  g
ms2
K
1

 2 2 1   Km
dz c p m s K
m
Concept: an ‘air parcel’
Z
p
It’s a useful
concept
to imagine
what will
happen as a
mass (‘parcel’)
of air moves up
or down in the
atmosphere.
Assumptions for a
parcel of air
- No exchange of mass with environment
- No exchange of heat with surrounding
- Adjusts to pressure of environment
(And moves slowly enough to neglect energy of
movement of air parcel)
Summary
• Air temperature generally decreases with
increasing altitude (e.g. radiosonde data)
• Using some physics (hydrostatic equation, 1st
Law of thermodynamics), we can derive a
theoretical expression for the temperature
gradient of an adiabatically ascending dry air
parcel: -9.8 K/km
• This is quite often a good approximation of the
real atmosphere
• Main complications involve moisture condensing
and releasing latent heat – next lecture.
Current Weather
Hugh Pumphrey’s web-pages:
https://www.geos.ed.ac.uk/homes/hcp/currentmet.html
Potential Temperature (θ)
• The potential temperature of an air parcel
is its temperature when compressed (or
expanded) adiabatically to surface
pressure (p0) (defined as a standard
pressure of 1000 hPa).
• Again, start from the 1st Law of
Thermodynamics, and make dq=0:
cpdT  dp  0
Ideal Gas Law (see Lecture 8)
p  RT
so:
1
RT
 

p
cpdT  dp  0
substitute in α:
RT
cpdT 
dp  0
p
Divide by RT:
cp dT dp

0
R T
p
Integrate both sides, from the starting (p,T) to
the surface (p0,T0), noting cp/R is a constant:
T
cp
dT


R T0   T
p
dp
p p
0
Remember integral of 1/x is natural log of x:
x2
1
x x dx 
1
ln x
x2
x1
 x2 
 ln x2   ln x1   ln 
 x1 
Integrating:
Remember:
Hence:
 p
cp  T 
ln   ln 
R  
 p0 
 
a(ln b)  ln ba
T 
 
 
cp
R
p

p0
or:
 p
  
  p0 
T
Rearrange to give potential temperature, θ:
 p0 
  T  
 p
R
cp
R
cp