Atmospheric Thermodynamics
Ben Harvey
NCAS, University of Reading
[email protected]
With thanks to: Alan Blyth, Maarten Ambaum
What is thermodynamics?
The study of heat and temperature and their relation to energy and work
• Individual molecules are ignored and the system is treated as a bulk entity
described by macroscopic variables such as temperature, pressure and
density
• Assumes the existence of equilibrium states of the system
Other approaches:
• Statistical mechanics – treat behaviour of molecules statistically
• Kinetic theory – use classical mechanics to describe the motions and
collisions of molecules
What is atmospheric thermodynamics?
Primitive equations:
𝐷𝑢
1 𝜕𝑝
− 𝑓𝑣 = −
+ 𝐹𝑥
𝐷𝑡
𝜌 𝜕𝑥
𝐷𝑣
1 𝜕𝑝
+ 𝑓𝑢 = −
+ 𝐹𝑦
𝐷𝑡
𝜌 𝜕𝑦
𝐷𝑤
1 𝜕𝑝
=𝑔−
+ 𝐹𝑧
𝐷𝑡
𝜌 𝜕𝑧
𝐷𝜌
+ 𝜌𝛻 ⋅ 𝒖 = 0
𝐷𝑡
𝐷𝜃
= ℎ𝑒𝑎𝑡𝑖𝑛𝑔
𝐷𝑡
Heating terms include:
• Radiative heating/cooling
• Latent heat exchange from
condensation/evaporation
• Conduction from surface
Only one equation, but diabatic
heating ultimately provides the
energy source for all weather systems
A broad range of topics useful for
atmospheric science, we’ll group
them loosely into two themes…
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
A typical atmospheric profile
THEME I:
The vertical structure of
the atmosphere
• How are temperature, pressure
and density related?
• What controls the structure of
the temperature profile?
• How fast can temperature
decrease with height before
the atmosphere is unstable?
Zonal mean temperature profile
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
The ideal gas law
Macroscopic variables describe the bulk properties of a gas
Volume
𝑉 [m3 ]
Mass
𝑀 = 𝑛𝜇 kg
(𝑛 = no. moles, 𝜇 = molar mass)
Temperature 𝑇 [K]
Density
𝜌 = 𝑀/𝑉 [kg m−3 ]
Pressure
𝑝 = force/area [Pa = Nm−2 ]
(often use the hectopascal: 1 hPa = 100 Pa)
𝜌
p M
T
V
The ideal gas law
Relationships between the macroscopic
variables can be determined by experiment or
theory – exact for ideal gases:
• Molecules assumed to have zero volume
• Molecules interact solely via elastic collisions
Assumptions are appropriate for low density
gases, and are a very good approximation for the
atmosphere (below ≈ 100 km)
The ideal gas law
Relationships between the macroscopic
variables can be determined by experiment or
theory – exact for ideal gases:
• Boyle’s Law
If temperature fixed:
• Charles’ Law
If pressure fixed:
Combine these to find:
What is the constant?
𝑝𝑉 = constant
𝑉∝𝑇
𝑝𝑉 ∝ 𝑇
The ideal gas law
• Ideal gas law
The constant of proportionality depends only on the number of molecules
present (but not the type of gas) – Avogadro’s law
So, for any gas:
More useful form:
𝑝𝑉 = 𝑛𝑅∗ 𝑇
𝑝 = 𝜌𝑅𝑇
where 𝑅 ∗ = 8.314 J K-1 mol-1
(the universal gas constant)
where 𝑅 = 𝑅∗ /𝜇
(the specific gas constant)
𝑝 = 𝜌𝑅𝑇
The ideal gas law
• Dalton’s law
For a mixture of gases: 𝑝 = 𝑖 𝑝𝑖 (𝑝𝑖 is the partial pressure of gas 𝑖)
 Can use the ideal gas law for mixtures of gases (e.g. air)
If gas 𝑖 has mass fraction 𝑐𝑖
then:
𝑅=
𝑐𝑖 𝑅𝑖
𝑖
For dry air:
𝑅 = 𝑅𝑑 = 287 J kg −1 K −1
The main constituents of air
For
moist air with
no condensate:
Constituent
Mass fraction
Molar mass
𝑅 = 1 + 0.61𝑞
𝑅]𝑑
𝜇 [g mol−1
𝑐𝑖 [g g −1 ]
where q is the
specific humidity
Nitrogen
28.02
0.755
Oxygen
32.00
0.231
Useful
trick
Argon
0.013
Use IGL with 39.93
𝑅 = 𝑅𝑑 but define
Carbon
44.01
~0.0006
thedioxide
virtual temperature
Water vapour
𝑇𝑣𝑖𝑟𝑡 = 118.02
+ 0.61𝑞 𝑇0 to ~0.03
Discussion I: pumping up car tyre
air
Specified
variables
Unknown
variables
pump tyre
𝑝 = 1 × 105 Pa
𝑇 = 288 K
𝜌 = 1.2 kg m−3
and
𝑝 = 𝜌𝑅𝑇
allow to cool
𝑝=?
𝑇 = 288 K
𝜌 = 2.4 kg m−3
Discussion I: pumping up car tyre
air
Specified
variables
Unknown
variables
pump tyre
𝑝 = 1 × 105 Pa
𝑇 = 288 K
𝜌 = 1.2 kg m−3
Ideal gas law is gives the pressure
and
𝑝 = 𝜌𝑅𝑇
allow to cool
𝑝 = 2 × 105 Pa
𝑇 = 288 K
𝜌 = 2.4 kg m−3
Discussion I: pumping up car tyre
air
Specified
variables
Unknown
variables
pump tyre
𝑝 = 1 × 105 Pa
𝑇 = 288 K
𝜌 = 1.2 kg m−3
𝑝 = 𝜌𝑅𝑇
allow to cool
𝑝=?
𝑇 =?
𝜌 = 2.4 kg m−3
Ideal gas law is 1 equation for three unknowns (𝑝, 𝜌, 𝑇)
Need more information to solve intermediate step

Next section: look at conservation of energy
𝑝 = 2 × 105 Pa
𝑇 = 288 K
𝜌 = 2.4 kg m−3
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
Heat and work
Work (W) is the energy transferred between bodies by mechanical forces.
𝐹
• The work done by a force 𝐹 is
d𝑊 = 𝐹d𝑥
• If the force acts to compress or expand
a gas this becomes
d𝑊 = −𝑝d𝑉
Piston
𝑑𝑥
𝑝 𝑉
Gas
Heat and work
Heat (Q) is the energy transferred between bodies of different temperature.
Evaporation
or
Trenberth et al (2009)
Heat and work
Heat can act to change temperature (sensible heat):
or phase (latent heat):
d𝑄 = 𝑀𝑐d𝑇
d𝑄 = 𝐿d𝑀
• For a gas, the specific heat capacity (c) depends on the setup:
𝑐𝑣 = heating at constant volume 𝑐𝑝 = heating at constant pressure
Which is larger?
• For an ideal gas:
• For dry air:
𝑐𝑝 = 𝑐𝑣 + 𝑅
𝑐𝑝 = 1004 J kg −1 K −1
Heat and work
Heat and work are transfers of energy between bodies.
Where does this energy go?
Answer: into an internal reservoir of energy called internal energy (U)
The first law of thermodynamics
Conservation of energy:
𝑑𝑈 = 𝑑𝑄 + 𝑑𝑊
(i.e. heat is a form of energy!)
For an ideal gas: 𝑈 = 𝑐𝑣 𝑇 (i.e. internal energy is just the KE of molecules)
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
𝑑𝑈 = 𝑑𝑄 + 𝑑𝑊
Adiabatic processes
A special class of processes have no heating (𝑑𝑄 = 0)
 called an adiabatic process
For gases, this just represents an expansion/compression without transferring
heat with surroundings (i.e. no radiation or condensation/evaporation)
For ideal gases, the first law becomes:
𝜅
which leads to:
𝑝
𝑇 = 𝑇0
𝑝0
where 𝑝0 and 𝑇0 are initial values, and 𝜅 =
𝑐𝑣 𝑑𝑇 = −𝑝𝑑𝑉
(the Poisson equation)
𝑅
𝑐𝑝
= 0.286 for dry air
Discussion II: pumping up car tyre
air
Specified
variables
Unknown
variables
pump tyre
allow to cool
dW only
(adiabatic)
dQ only
𝑝 = 1 × 105 Pa
𝑇 = 288 K
𝜌 = 1.2 kgm−3
𝑝 = 2.6 × 105 Pa
𝑇 = 380 K
𝜌 = 2.4 kgm−3
𝑑𝑈 = 𝑑𝑄 + 𝑑𝑊
𝑝 = 2 × 105 Pa
𝑇 = 288 K
𝜌 = 2.4 kgm−3
Assuming pumping is adiabatic (no heat lost to surroundings).
IGL law plus Poisson equation solves the intermediate step.
Adiabatic processes
Let’s apply to the atmosphere!
Consider moving a parcel of air (𝑇,𝑝) down to the surface adiabatically
Its final temperature is called the potential temperature:
1000 hPa
𝜃=𝑇
𝑝
𝜅
• It is a property of air parcels that is approximately conserved in time
• Modified only by diabatic processes:
𝐷𝜃
𝐷𝑡
=
𝑝0 𝜅 𝑞
𝑝
𝑐𝑝
• It is also important for stability – see next section
(i.e. heating)
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
Buoyancy and static stability
Level 2
Level 1
1000 hPa
𝜃=𝑇
𝑝
Consider lifting a parcel adiabatically
from level 1 to level 2:
𝑝2 , 𝜌𝑃 , 𝑇𝑃 , 𝜃1
• By Archimedes, the net buoyancy
on the parcel:
𝑝2 , 𝜌2 , 𝑇2 , 𝜃2 force per unit mass
𝜌2 − 𝜌𝑃
𝐵=𝑔
𝜌𝑃
𝑝1 , 𝜌1 , 𝑇1 , 𝜃1 • But 𝜌𝑃 is related to 𝜃1 , find:
𝜃1 − 𝜃2
𝐵=𝑔
𝜃2
Parcel will rise further if 𝜃1 > 𝜃2  atmosphere unstable to convection
Parcel will not accelerate if 𝜃1 = 𝜃2  neutral
Parcel will sink towards original location if 𝜃1 < 𝜃2  stable
𝜅
Buoyancy and static stability
In summary: a (dry) atmosphere is
Stable
𝑑𝜃
>0
𝑑𝑧
Neutral
𝑑𝜃
=0
𝑑𝑧
Unstable
𝑑𝜃
<0
𝑑𝑧
Buoyancy and static stability
Gravity waves:
• Generated by flow over
orography
• Also by: flow over atmospheric
fronts, convective clouds,
instabilities of the jet stream, …
• Frequencies
≤𝑁=
𝑔 𝑑𝜃
𝜃 𝑑𝑧
1
2
≈ 1/(5 min)
(Brunt-Vaisala frequency)
Buoyancy and static stability
In summary: a (dry) atmosphere is
Stable
𝑑𝜃
>0
𝑑𝑧
Neutral
𝑑𝜃
=0
𝑑𝑧
Unstable
𝑑𝜃
<0
𝑑𝑧
Buoyancy and static stability
Special case:
d𝜃
𝑑𝑧
=0
A neutrally-stable / well-mixed dry atmosphere
Find: 𝑇(𝑧) = 𝑇𝑠 − Γ𝑑 𝑧 (use hydrostatic balance)
where Γ𝑑 =
𝑔
𝑐𝑝
≈ 10 K km−1 is the dry adiabatic lapse rate
 𝑇 can decrease with height and still be stable, but not faster than Γ𝑑
Buoyancy and static stability
In summary: a (dry) atmosphere is
Stable
𝑑𝜃
>0
𝑑𝑧
𝑑𝑇
> −Γ𝑑
𝑑𝑧
Neutral
𝑑𝜃
=0
𝑑𝑧
𝑑𝑇
= −Γ𝑑
𝑑𝑧
Unstable
𝑑𝜃
<0
𝑑𝑧
𝑑𝑇
< −Γ𝑑
𝑑𝑧
(Γ𝑑 = 10 K km−1 )
Recap of THEME I
• Pressure, density and temperature
are related via the ideal gas law
• Energy is transferred with the
surroundings via heat and work
• In the absence of heating, air
parcels conserve their potential
temperature
• Potential temperature also
determines the stability of the
atmosphere
So, what sets the
temperature profile?
Above the tropopause:
Radiative heating dominates
Below the tropopause:
The radiative equilibrium profile
is unstable (dashed line)
 convection and weather
systems mix the air
 result is a T profile close to
neutral stability (but moisture
modifies this – see next section)
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
THEME II:
The special role of
moisture
• Water is the only substance to
commonly occur in all three
phases in the atmosphere
• Latent heat is released into the
atmosphere when clouds form
• The transport of water vapour is
an important contributor to both
the global energy budget and
the development of individual
weather systems
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
Latent heat
Heat acting to change phase
Of an amount of substance d𝑀:
d𝑄 = 𝐿 d𝑀
where 𝐿 is either:
the latent heat of fusion (𝐿𝑓 ), or
the latent heat of vaporization (𝐿𝑣 )
Latent heat
For water:
𝐿𝑓 = 3.3 × 105 J kg −1
𝐿𝑣 = 25 × 105 J kg −1
Compare with specific heat capacity
of liquid water:
𝑐 = 4.2 × 103 J kg −1 K −1
 Water vapour is responsible for
large amounts of energy transport in
the atmosphere
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
Saturation
Consider a closed box containing liquid water
and water vapour at pressure 𝑒.
There is a continual exchange of molecules
between the vapour and liquid phases.
How much water vapour is there at
equilibrium?
The first law tells us indirectly. It gives:
𝑑𝑒𝑠
𝑑𝑇
=
𝐿𝑣
𝑇(𝜌𝑣−1 −𝜌𝑙−1 )
(Clausius-Clapeyron eqn)
where 𝑒𝑠 (𝑇) is the saturation vapour pressure.
Saturation
Need to integrate to find 𝑒𝑠 𝑇 .
Easiest to use an empirical table or
formula, e.g. Teten’s formula:
16.67 𝑇[deg 𝐶]
𝑒𝑠 𝑇 ≈ 6.112 exp
243.5 + 𝑇[deg 𝐶]
The saturation vapour pressure increases
strongly with temperature
• An increase of 10C increases 𝑒𝑠 by 7%
• An increase of 100C doubles 𝑒𝑠
Saturation
The saturation vapour pressure is
smaller for ice than for water:
Ice crystals can grow at the
expense of water droplets (known
as the Bergeron-Findeisen process)
An aside on (some of the many) moisture variables…
How to quantify how much water vapour is in the air?
• Vapour pressure:
𝑒
• Mass mixing ratio:
𝜌𝑣
𝑟=
𝜌𝑑
• Specific humidity:
𝜌𝑣
𝑒
𝑞=
= 1.61
𝜌𝑑 + 𝜌𝑣
𝑝
• Saturation vapour pressure:
𝑒𝑠 (𝑇)
• Saturation mass mixing ratio:
𝜌𝑣,𝑠 (𝑇)
𝑟𝑠 =
𝜌𝑑
• Relative humidity:
𝑒
𝑅𝐻 % = 100
𝑒𝑠 (𝑇)
An aside on (some of the many) moisture variables…
• Dew point temperature:
𝑒𝑠 𝑇𝑑 = 𝑒
“The temperature to which air must
be cooled isobarically to achieve
saturation”
• Wet bulb temperature:
𝑇𝑤 (get from psychrometric table)
“The temperature to which air can be
cooled isobarically by evaporating
water into it”
e [hPa]
𝑇𝑤
𝑇𝑑 𝑇
A whirling psychrometer
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
Stability with moisture
Recall from earlier:
If we lift a dry air parcel through
a well-mixed dry atmosphere it
will expand and cool at the DALR
(≈ 10 K km−1 )
Now suppose there is moisture in
the parcel:
Condensation will occur when
the RH reaches 100%, releasing
latent heat which partiallyoffsets the cooling
Stability with moisture
Again we can find the change in
temperature with height from the
first law:
𝑐𝑣 d𝑇 = 𝐿d𝑟𝑠 − 𝑝d𝑉
Solve to find the moist adiabatic
lapse rate:
𝑑𝑇
≈ 𝑇𝑠 − Γ𝑠 𝑧
𝑑𝑧
with Γ𝑠 ≈ 5 K km−1
Stability with moisture
The equivalent potential temperature
is conserved for moist adiabatic
ascent:
𝐿𝑟𝑠
𝜃𝑒 ≈ 𝜃exp
𝑐𝑝,𝑑 𝑇
It is the potential temperature the
parcel would have if all the water
vapour was condensed out.
To find: lift saturated parcel to high
altitude along its moist adiabat, then
return to surface along a dry adiabat.
𝜃
𝜃𝑒
Buoyancy and static stability
In summary: a moist atmosphere is
Stable
𝑑𝜃𝑒
>0
𝑑𝑧
𝑑𝑇
> −Γ𝑠
𝑑𝑧
Neutral
𝑑𝜃𝑒
=0
𝑑𝑧
𝑑𝑇
= −Γ𝑠
𝑑𝑧
Unstable
𝑑𝜃𝑒
<0
𝑑𝑧
𝑑𝑇
< −Γ𝑠
𝑑𝑧
(Γ𝑠 ≈ 5 K km−1 )
Outline of this session
1.
2.
3.
4.
THEME I:
The vertical structure of the
atmosphere
The ideal gas law
Heat and work
Adiabatic processes
Buoyancy and static stability
THEME II:
The special role of moisture
5.
6.
7.
8.
Latent heat
Saturation
Stability with moisture
The tephigram
A useful tool for graphically
calculating many of the
variables discussed today
from a balloon sounding
Pressure [hPa]
The tephigram
Temperature [deg C]
The tephigram
Plot profiles of temperature
and dewpoint temperature
from a balloon ascent
Td
T
Pressure [hPa]
Lines indicate:
• Pressure
• Temperature
Temperature [deg C]
Lines indicate:
• Pressure
• Temperature
Other lines indicate:
• Dry adiabat (constant 𝜃)
• Moist adiabat (constant 𝜃𝑒 )
• Saturated mixing ratio
(constant 𝑟𝑠 )
Pressure [hPa]
The tephigram
Normand construction
𝑇𝑑 𝑇𝑤 𝑇
Temperature [deg C]
Tephigram exercise
12Z on 12th November
Storm Abigail – 12/13 November 2015
00Z on 13th November
UK Met Office
Dundee Satellite Receiving Station
Tephigram exercise
1. Identify the pressure and
temperature lines on your
tephigram
2. Plot the temperature
profile
3. Plot the dewpoint
temperature profile
4. Interpret the features of
the airmasses from your
plot
Tephigram exercise
Recap of THEME II:
• Latent heat release in clouds is an
important process
• The amount of water vapour that
air can hold depends only on
temperature (Clausius-Clapyron
equation)
• Equivalent potential temperature
is conserved during phase
changes and determines stability
• The tephigram allows a quick
assessment of atmospheric
soundings
The end!
Any questions?
Extra slides…
Hydrostatic balance
𝑝 + Δ𝑝
Consider the forces acting on a parcel of air at rest:
𝑝𝐴 = 𝑝 + Δ𝑝 𝐴 + 𝑀𝑔
But 𝑀 = 𝜌𝑉 = 𝜌𝐴Δ𝑧, so:
𝑑𝑝
= −𝜌𝑔
𝑑𝑧
𝐴
𝑀𝑔
Hydrostatic balance
This is a good approximation in the atmosphere
(except where vertical accelerations are large, e.g.
inside convective storms, at strong fronts, etc)
𝑝
Δ𝑧
𝑑𝑝
= −𝜌𝑔
𝑑𝑧
Hydrostatic balance
Implications:
• Air pressure is due solely to the weight of air above
• Pressure decreases with height. By ideal gas law: −𝜌𝑔 = −𝑝𝑔/𝑅𝑇

pressure decreases faster in colder (denser) air
• If 𝑇 is constant, can integrate to find:
𝑝 𝑧 = 𝑝𝑠 𝑒
• Which leads to:
𝑧=
−
𝑔𝑧
𝑅𝑇
𝑅𝑇
log(𝑝𝑠 /𝑝(𝑧))
𝑔
(exponential decay with height)
(the altimeter equation)
Discussion: hydrostatic balance
𝑑𝑝
= −𝜌𝑔
𝑑𝑧
• What is the total weight of air in an atmospheric column?
• At what depth in the ocean is pressure equal to double atmospheric
pressure?
Unsaturated moist air
Unsaturated moist air is simply a mixture of gases
Total system:
𝑝 = 𝜌𝑅𝑚 𝑇
where 𝑅𝑚 = 𝑅𝑑 (1 + 0.61𝑞)
and 𝑞 = 𝜌𝑣 /𝜌 is the specific humidity
Dry air
𝑝 − 𝑒 = 𝜌𝑑 𝑅𝑑 𝑇
Water vapour
𝑒 = 𝜌𝑣 𝑅𝑣 𝑇
Convenient to write: 𝑝 = 𝜌𝑅𝑑 𝑇𝑣𝑖𝑟𝑡 where 𝑇𝑣𝑖𝑟𝑡 = 𝑇(1 + 0.61𝑞)
then all of dry results carry over to moist air with only minor corrections:
𝑐𝑝 = 𝑐𝑝,𝑑 (1 + 0.87𝑞), Γ𝑎𝑑𝑖𝑎𝑏 = Γ𝑑 /(1 + 0.87𝑞)
stability depends on 𝑑𝜃𝑣𝑖𝑟𝑡 /𝑑𝑧