Atmospheric Thermodynamics Ben Harvey NCAS, University of Reading b.j.harvey@reading.ac.uk 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: 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… 𝐷𝑢 𝐷𝑡 −𝑓𝑣=− 1 𝜌 𝜕𝑝 𝜕𝑥 + 𝐹 𝑥 𝐷𝑣 𝐷𝑡 +𝑓𝑢=− 1 𝜌 𝜕𝑝 𝜕𝑦 + 𝐹 𝑦 𝐷𝑤 𝐷𝑡 =𝑔− 1 𝜌 𝜕𝑝 𝜕𝑧 + 𝐹 𝑧 𝐷𝜌 𝐷𝑡 +𝜌𝛻⋅𝒖=0 𝐷𝜃 𝐷𝑡 =ℎ𝑒𝑎𝑡𝑖𝑛𝑔

Outline of this session 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 Latent heat Saturation Stability with moisture The tephigram

THEME I: The vertical structure of the atmosphere 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 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 Latent heat Saturation Stability with moisture The tephigram

The ideal gas law Macroscopic variables describe the bulk properties of a gas Volume 𝑉 [ m 3 ] 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=N m −2 ] (often use the hectopascal: 1 hPa=100 Pa) p M T V 𝜌

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: 𝑝𝑉= constant Charles’ Law If pressure fixed: 𝑉∝𝑇 Combine these to find: 𝑝𝑉∝𝑇 What is the constant?

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: where 𝑅 ∗ = 8.314 J K-1 mol-1 (the universal gas constant) More useful form: where 𝑅= 𝑅 ∗ /𝜇 (the specific gas constant) 𝑝𝑉=𝑛 𝑅 ∗ 𝑇 𝑝=𝜌𝑅𝑇

The main constituents of air 𝑝=𝜌𝑅𝑇 The ideal gas law Dalton’s law For a mixture of gases: 𝑝= 𝑖 𝑝 𝑖 ( 𝑝 𝑖 is the partial pressure of gas 𝑖) Can use the ideal gas law for mixtures of gases (e.g. air) The main constituents of air If gas 𝑖 has mass fraction 𝑐 𝑖 then: 𝑅= 𝑖 𝑐 𝑖 𝑅 𝑖 For dry air: 𝑅= 𝑅 𝑑 =287 J k g −1 K −1 For moist air with no condensate: 𝑅= 1+0.61𝑞 𝑅 𝑑 where q is the specific humidity Useful trick Use IGL with 𝑅= 𝑅 𝑑 but define the virtual temperature 𝑇 𝑣𝑖𝑟𝑡 = 1+0.61𝑞 𝑇 Constituent Molar mass 𝜇 [g mo l −1 ] Mass fraction 𝑐 𝑖 [g g −1 ] Nitrogen 28.02 0.755 Oxygen 32.00 0.231 Argon 39.93 0.013 Carbon dioxide 44.01 ~0.0006 Water vapour 18.02 0 to ~0.03

Discussion I: pumping up car tyre 𝑝=𝜌𝑅𝑇 air pump tyre and allow to cool Specified variables Unknown variables 𝑝=1× 10 5 Pa 𝑇=288 K 𝜌=1.2 kg m −3 𝑝= ? 𝑇=288 K 𝜌=2.4 kg m −3

Discussion I: pumping up car tyre 𝑝=𝜌𝑅𝑇 air pump tyre and allow to cool Specified variables Unknown variables 𝑝=1× 10 5 Pa 𝑇=288 K 𝜌=1.2 kg m −3 𝑝=2× 10 5 Pa 𝑇=288 K 𝜌=2.4 kg m −3 Ideal gas law is gives the pressure

Discussion I: pumping up car tyre 𝑝=𝜌𝑅𝑇 air pump tyre allow to cool Specified variables Unknown variables 𝑝=1× 10 5 Pa 𝑇=288 K 𝜌=1.2 kg m −3 𝑝= ? 𝑇= ? 𝜌=2.4 kg m −3 𝑝=2× 10 5 Pa 𝑇=288 K 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

Outline of this session 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 Latent heat Saturation Stability with moisture The tephigram

Heat and work 𝐹 Piston 𝑑𝑥 𝑝 𝑉 Gas Work (W) is the energy transferred between bodies by mechanical forces. 𝐹 𝑝 𝑉 𝑑𝑥 Gas Piston The work done by a force 𝐹 is d𝑊=𝐹d𝑥 If the force acts to compress or expand a gas this becomes d𝑊=−𝑝d𝑉

Heat and work Heat (Q) is the energy transferred between bodies of different temperature. Evaporation or Trenberth et al (2009)

𝑐 𝑣 = heating at constant volume 𝑐 𝑝 = heating at constant pressure Heat and work Heat can act to change temperature (sensible heat): d𝑄=𝑀𝑐d𝑇 or phase (latent heat): d𝑄=𝐿d𝑀 For a gas, the specific heat capacity (c) depends on the setup: 𝑐 𝑣 = heating at constant volume 𝑐 𝑝 = heating at constant pressure Which is larger? For an ideal gas: 𝑐 𝑝 = 𝑐 𝑣 +𝑅 For dry air: 𝑐 𝑝 =1004 J k g −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: (i.e. heat is a form of energy!) For an ideal gas: 𝑈= 𝑐 𝑣 𝑇 (i.e. internal energy is just the KE of molecules) 𝑑𝑈=𝑑𝑄+𝑑𝑊

Outline of this session 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 Latent heat Saturation Stability with moisture The tephigram

Adiabatic processes 𝑑𝑈=𝑑𝑄+𝑑𝑊 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: 𝑐 𝑣 𝑑𝑇=−𝑝𝑑𝑉 which leads to: (the Poisson equation) where 𝑝 0 and 𝑇 0 are initial values, and 𝜅= 𝑅 𝑐 𝑝 =0.286 for dry air 𝑇= 𝑇 0 𝑝 𝑝 0 𝜅

Discussion II: pumping up car tyre 𝑑𝑈=𝑑𝑄+𝑑𝑊 air pump tyre allow to cool dW only (adiabatic) dQ only Specified variables Unknown variables 𝑝=1× 10 5 Pa 𝑇=288 K 𝜌=1.2 kg m −3 𝑝=2.6× 10 5 Pa 𝑇=380 K 𝜌=2.4 kg m −3 𝑝=2× 10 5 Pa 𝑇=288 K 𝜌=2.4 kg m −3 𝑝=2× 10 5 Pa 𝑇=288 K 𝜌=2.4 kg m −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: It is a property of air parcels that is approximately conserved in time Modified only by diabatic processes: 𝐷𝜃 𝐷𝑡 = 𝑝 0 𝑝 𝜅 𝑞 𝑐 𝑝 (i.e. heating) It is also important for stability – see next section 𝜃=𝑇 1000 hPa 𝑝 𝜅

Outline of this session 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 Latent heat Saturation Stability with moisture The tephigram

Buoyancy and static stability 𝜃=𝑇 1000 hPa 𝑝 𝜅 Buoyancy and static stability Consider lifting a parcel adiabatically from level 1 to level 2: By Archimedes, the net buoyancy force per unit mass on the parcel: 𝐵=𝑔 𝜌 2 − 𝜌 𝑃 𝜌 𝑃 But 𝜌 𝑃 is related to 𝜃 1 , find: 𝐵=𝑔 𝜃 1 − 𝜃 2 𝜃 2 𝑝 1 , 𝜌 1 , 𝑇 1 , 𝜃 1 𝑝 2 , 𝜌 2 , 𝑇 2 , 𝜃 2 𝑝 2 , 𝜌 𝑃 , 𝑇 𝑃 , 𝜃 1 Level 2 Level 1 Parcel will rise further if 𝜃 1 > 𝜃 2  atmosphere unstable to convection Parcel will not accelerate if 𝜃 1 = 𝜃 2  neutral Parcel will sink towards original location if 𝜃 1 < 𝜃 2  stable

Buoyancy and static stability In summary: a (dry) atmosphere is Stable 𝑑𝜃 𝑑𝑧 >0 Neutral 𝑑𝜃 𝑑𝑧 =0 Unstable 𝑑𝜃 𝑑𝑧 <0

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 𝜃 𝑑𝜃 𝑑𝑧 1 2 ≈1/(5 min) (Brunt-Vaisala frequency)

Buoyancy and static stability In summary: a (dry) atmosphere is Stable 𝑑𝜃 𝑑𝑧 >0 Neutral 𝑑𝜃 𝑑𝑧 =0 Unstable 𝑑𝜃 𝑑𝑧 <0

Buoyancy and static stability Special case: d𝜃 𝑑𝑧 =0 A neutrally-stable / well-mixed dry atmosphere Find: 𝑇(𝑧)= 𝑇 𝑠 − Γ 𝑑 𝑧 (use hydrostatic balance) where Γ 𝑑 = 𝑔 𝑐 𝑝 ≈10 K k m −1 is the dry adiabatic lapse rate  𝑇 can decrease with height and still be stable, but not faster than Γ 𝑑

Buoyancy and static stability In summary: a (dry) atmosphere is Stable 𝑑𝜃 𝑑𝑧 >0 𝑑𝑇 𝑑𝑧 > −Γ 𝑑 Neutral 𝑑𝜃 𝑑𝑧 =0 𝑑𝑇 𝑑𝑧 = −Γ 𝑑 Unstable 𝑑𝜃 𝑑𝑧 <0 𝑑𝑇 𝑑𝑧 < −Γ 𝑑 ( Γ 𝑑 =10 K k m −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 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 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 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 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 ( 𝐿 𝑓 ), or the latent heat of vaporization ( 𝐿 𝑣 )

Latent heat For water: 𝐿 𝑓 =3.3× 10 5 J k g −1 𝐿 𝑣 =25× 10 5 J k g −1 Compare with specific heat capacity of liquid water: 𝑐=4.2× 10 3 J k g −1 K −1  Water vapour is responsible for large amounts of energy transport in the atmosphere

Outline of this session 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 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: 𝑑 𝑒 𝑠 𝑑𝑇 = 𝐿 𝑣 𝑇( 𝜌 𝑣 −1 − 𝜌 𝑙 −1 ) (Clausius-Clapeyron eqn) where 𝑒 𝑠 (𝑇) is the saturation vapour pressure.

𝑒 𝑠 𝑇 ≈6.112 exp 16.67 𝑇[ deg 𝐶 ] 243.5+𝑇[ deg 𝐶 ] Saturation Need to integrate to find 𝑒 𝑠 𝑇 . Easiest to use an empirical table or formula, e.g. Teten’s formula: 𝑒 𝑠 𝑇 ≈6.112 exp 16.67 𝑇[ deg 𝐶 ] 243.5+𝑇[ deg 𝐶 ] The saturation vapour pressure increases strongly with temperature An increase of 10C increases 𝑒 𝑠 by 7% An increase of 100C doubles 𝑒 𝑠

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: 𝑟= 𝜌 𝑣 𝜌 𝑑 Specific humidity: 𝑞= 𝜌 𝑣 𝜌 𝑑 + 𝜌 𝑣 =1.61 𝑒 𝑝 Saturation vapour pressure: 𝑒 𝑠 (𝑇) Saturation mass mixing ratio: 𝑟 𝑠 = 𝜌 𝑣,𝑠 (𝑇) 𝜌 𝑑 Relative humidity: 𝑅𝐻 % =100 𝑒 𝑒 𝑠 (𝑇)

An aside on (some of the many) moisture variables… Dew point temperature: 𝑒 𝑠 𝑇 𝑑 =𝑒 “The temperature to which air must be cooled isobarically to achieve saturation” Wet bulb temperature: 𝑇 𝑤 (get from psychrometric table) “The temperature to which air can be cooled isobarically by evaporating water into it” e [hPa] 𝑇 𝑑 𝑇 𝑇 𝑤 A whirling psychrometer

Outline of this session 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 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 k m −1 ) Now suppose there is moisture in the parcel: Condensation will occur when the RH reaches 100%, releasing latent heat which partially- offsets the cooling

Stability with moisture Again we can find the change in temperature with height from the first law: 𝑐 𝑣 d𝑇=𝐿d 𝑟 𝑠 −𝑝d𝑉 Solve to find the moist adiabatic lapse rate: 𝑑𝑇 𝑑𝑧 ≈ 𝑇 𝑠 − Γ 𝑠 𝑧 with Γ 𝑠 ≈5 K k m −1

Stability with moisture The equivalent potential temperature is conserved for moist adiabatic ascent: 𝜃 𝑒 ≈𝜃exp⁡ 𝐿 𝑟 𝑠 𝑐 𝑝,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 𝑒 𝜃

Buoyancy and static stability In summary: a moist atmosphere is Stable 𝑑 𝜃 𝑒 𝑑𝑧 >0 𝑑𝑇 𝑑𝑧 > −Γ 𝑠 Neutral 𝑑 𝜃 𝑒 𝑑𝑧 =0 𝑑𝑇 𝑑𝑧 = −Γ 𝑠 Unstable 𝑑 𝜃 𝑒 𝑑𝑧 <0 𝑑𝑇 𝑑𝑧 < −Γ 𝑠 ( Γ 𝑠 ≈5 K k m −1 )

Outline of this session 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 Latent heat Saturation Stability with moisture The tephigram

The tephigram A useful tool for graphically calculating many of the variables discussed today from a balloon sounding Pressure [hPa] Temperature [deg C]

The tephigram T Td Lines indicate: Pressure Temperature Plot profiles of temperature and dewpoint temperature from a balloon ascent Pressure [hPa] Temperature [deg C]

The tephigram 𝑇 𝑑 𝑇 𝑤 𝑇 Lines indicate: Pressure Temperature Other lines indicate: Dry adiabat (constant 𝜃) Moist adiabat (constant 𝜃 𝑒 ) Saturated mixing ratio (constant 𝑟 𝑠 ) Normand construction Pressure [hPa] 𝑇 𝑑 𝑇 𝑤 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 Identify the pressure and temperature lines on your tephigram Plot the temperature profile Plot the dewpoint temperature profile 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 𝐴 Δ𝑧 𝑀𝑔 𝑝 Consider the forces acting on a parcel of air at rest: 𝑝𝐴= 𝑝+Δ𝑝 𝐴+𝑀𝑔 But 𝑀=𝜌𝑉=𝜌𝐴Δ𝑧, so: 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) 𝑑𝑝 𝑑𝑧 =−𝜌𝑔

Hydrostatic balance 𝑑𝑝 𝑑𝑧 =−𝜌𝑔 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 = 𝑝 𝑠 𝑒 − 𝑔𝑧 𝑅𝑇 (exponential decay with height) Which leads to: 𝑧= 𝑅𝑇 𝑔 log ( 𝑝 𝑠 /𝑝(𝑧)) (the altimeter equation)

Discussion: hydrostatic balance 𝑑𝑝 𝑑𝑧 =−𝜌𝑔 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=𝜌 𝑅 𝑚 𝑇 where 𝑅 𝑚 = 𝑅 𝑑 (1+0.61𝑞) and 𝑞= 𝜌 𝑣 /𝜌 is the specific humidity Convenient to write: 𝑝=𝜌 𝑅 𝑑 𝑇 𝑣𝑖𝑟𝑡 where 𝑇 𝑣𝑖𝑟𝑡 =𝑇(1+0.61𝑞) then all of dry results carry over to moist air with only minor corrections: 𝑐 𝑝 = 𝑐 𝑝,𝑑 (1+0.87𝑞), Γ 𝑎𝑑𝑖𝑎𝑏 = Γ 𝑑 /(1+0.87𝑞) stability depends on 𝑑𝜃 𝑣𝑖𝑟𝑡 /𝑑𝑧 Dry air 𝑝−𝑒 = 𝜌 𝑑 𝑅 𝑑 𝑇 Water vapour 𝑒= 𝜌 𝑣 𝑅 𝑣 𝑇