Numerical Weather Prediction Parametrization of diabatic processes Introduction to Moist Processes

1. ADRIAN Numerical Weather Prediction Parametrization of diabatic processesIntroduction to Moist Processes Adrian Tompkins tompkins@ecmwf.int

2. Clouds Water in the atmosphere can be present in all three phases: vapour, liquid and solid Clouds, water in the form of liquid or ice, play an crucial role in the earth’s climate Clouds and the precipitation formed in them are important forecast products

3. Clouds - why are they important? Clouds are a common atmospheric feature and cover large parts of the globe What is the mean coverage? Latent heat release or consumption occur either directly in clouds or in the precipitation produced in them.

4. Clouds - Why are they important? The entire model hydrological cycle will depend on the representation of clouds and their microphysical processes

5. They strongly affect the radiative fluxes throughout the atmosphere Sherwood et al JGR 1994, show that the tropical circulation is strongly influenced by the interaction between radiation and clouds Clouds - Why are they important? Cloud radiative effects (“forcing”) TOA- ERBE - JJA 87

6. GCM Grid cell 20-400km T,q,u,v,w... Clouds and Convection in GCMs - What are the problems ? Many clouds and especially the processes within them are of subgrid-scale size (both horizontally and vertically) and thus must be parameterized. This means a mathematical model is constructed that attempts to assess their effects in terms of the Large-scale model resolved quantities. The aim of this course component is to introduce a few of the approaches used for this important task

7. Clouds in GCMs - What are the problems ? There is a huge variety of cloud types stratocumulus cirrus cumulus/cumulonimbus http://australiasevereweather.com/photography

8. In this course, we will consider: composite infra-red image Warm and ice phase microphysics

9. This schematic of a front illustrates the interactions involved

10. In this course, we will consider: Stratocumulus Why are they dark?

11. Clouds in GCMs - What are the problems ? Stratocumulus Driven by turbulence in the boundary layer Very common, particularly over the oceans near the west Coast of the major continents Difficult to model: formation, regime (cloud cover), decoupling, break up… one of the common deficiencies of climate and forecast models

12. zi zcb zi Stratocumulus dry BL Shallow cumulus Deep cumulus Schematic: StCu to Cu Transition • destabilization: • surface buoyancy flux • atmos. radiative cooling • uplift (cooling, moistening) • convergence • shear Figure from Martin Koehler

13. JJA DJF Cloud Cover ECMWF ISCCP D2 Difference

14. In this course, we will consider: Cirrus

15. Clouds in GCMs - What are the problems ? Cirrus Ice cloud found in the upper troposphere. e.g: fronts, convective anvils. Important for the radiative impact Microphysics of the ice phase even less well understood than liquid phase

16. In this course, we will consider: Deep and Shallow (un)organised convection

17. Convection In atmospheric science generally restrict the term “convection” to refer to fluid motions resulting from the action of gravity on unstable vertical mass distribution. • Encompasses a huge range of scales of organisation: • Turbulent boundary Layers • Deep moist convection - Thunder storms • Squall Lines and Hurricanes Current models do not take organised systems into account

18. Complex interaction of a host of processes

19. Overview of Clouds/Convection • Introduction • Motivation, moist thermodynamics • Parametrization of moist convection(4 lectures, Peter), • Theory of moist convection • Common approaches to parametrization including the ECMWF scheme • Cloud Resolving Models(1 lecture - Adrian) • their use as parametrization tools • Parametrization of clouds(3 lectures - Adrian) • Basic microphysics of clouds • The problem of representing cloud cover • The ECMWF cloud scheme (and issues concerning validation) • Exercise Classes (1 afternoon, Peter and Adrian)

20. Moist Thermodynamics Brief Overview of what we presume you are familiar with, and an introduction to thermodynamic charts For simplified Overview: Rogers and Yau (1989) “A short course in cloud physics” For rigorous definitions: K. A. Emanuel (1994) “Atmospheric Convection” How many have used tephigrams?

21. Density=1/specific volume Moist Thermodynamics • Assume that “moist air” can be treated as mixture of two ideal gases: “dry air” + vapour Dry air equation of state: Temperature Gas Constant for dry air = 287 J Kg-1 K-1 Pressure Water Vapour equation of state: Vapour pressure vapour density Gas constant for Vapour = 461 J Kg-1 K-1 0.622

22. Heat is a form of energy, and energy is conserved (lowercase=per unit mass) Heat supplied by diabatic process Change in internal Energy Work done by Gas Can write as Specific Heat at constant volume Specific Heat at constant pressure First law of thermodynamics

23. Differentiating the equation of state and using Special processes: “Adiabatic Process” dq=0 Special significance since many atmospheric motions can be approximated as adiabatic First law of thermodynamics: form

24. Setting reference pressure to 1000hPa gives the definition of potential temperature (Note: for dry air) Using equation of state and integrating, obtain Poisson’s equation Conserved Variables

25. rotate to have pressure (almost) horizontal Thus diagram with ordinates T versus ln qwill have the properties of “equal areas”=“equal energy” Called a TEPHIGRAM Meteorological energy diagrams Total heat added in cyclic process:

26. Consider this closed system in equilibrium:T equal for water & air, no net evaporation or condensation Air is said to be saturated air+water vapour water • For the phase change between water and water vapour the equilibrium pressure (often called saturation water vapour pressure) is a function of temperature only The Clausius-Clapeyron equation • with av >> aw, and the ideal gas lawav=RvT/es

27. The Clausius-Clapeyron equation - Integration • The problem of integrating the Clausius-Clapeyron equation lies in the temperature dependence of Lv. • Fortunately this dependence is only weak, so that approximate formulae can be derived. Nonlinearity has consequences for mixing in convection es0 = 6.11 hPa at T0=273 K

28. Humidity variablesThere are a number of common ways to describe vapour content : 1. Vapour Pressure, e 2. Absolute humidity 3. Specific humidity Mass of water vapour per moist air mass 4. Mixing ratio 5. Relative humidity

29. T q r pressure Tephigram (II) Saturation specific humidity Saturation mixing ratio Function of temperature and pressure only – tephigrams have isopleths of rs

30. Using a Tephigram At a pressure of 950 hPa Measure T=20 oC r=10 gkg-1 plot a atmospheric sounding

31. Another way to describe the vapour content is the virtual temperature , an artificial temperature. By extension, we define the virtual potential temperature, which is a conserved variable in unsaturated ascent, and related to density Virtual temperatureTv It describes the temperature dry air would have to have in order to have the same density as a sample of moist air

32. absolute liquid water content • liquid water mixing ratio • total water mixing ratio • specific liquid water content Analogous toTvcan define the density temperature Tr, which is the T dry air would have to have equal density to moist cloudy air Water variablesThe description of water content in its liquid (and/or ice) state How can we saturate a parcel of air?

33. Temperature to which air must be cooled to reach saturation, with p and r held constant Cooling:Dew point temperatureTd Evaporation: Wet-bulb temperature Tw Temperature to which air may be cooled by evaporation of water into it until saturation is reached, at constant p Will show how to determine graphically from tephigram Ways of reaching saturation • Several ways to reach saturation: • Diabatic Cooling (e.g. Radiation) • Evaporation (e.g. of precipitation) • Expansion (e.g. ascent/descent) All of these are important cloud processes!!!

34. Ways of reaching saturation:Expansion: (Pseudo) Adiabatic Processes As (unsaturated) moist air expands (e.g. through vertical motion), cools adiabatically conserving q. Eventually saturation pressure is reached, T,p are known as the “isentropic condensation temperature and pressure”, respectively. If expansion continues, condensation will occur (assuming that liquid water condenses efficiently and no super saturation can persist), thus the temperature will decrease at a slower rate.

35. Ways of reaching saturation:Expansion: (Pseudo) Adiabatic Processes • Have to make a decision concerning the condensed water. • Does it falls out instantly or does it remain in the parcel? If it remains, the heat capacity should be accounted for, and it will have an effect on parcel buoyancy • Once the freezing point is reached, are ice processes taken into account? (complex) • These are issues concerning microphysics, and dynamics. The air parcel history will depend on the situation. We take the simplest case: all condensate instantly lost as precipitation, known as “Pseudo adiabatic process” Pseudo adiabatic process

36. T q r For a pseudoadiabatic process: gives: pressure Pseudoadiabat (or moist adiabat) Tephigram (III) Remember: Involves an arbitrary “cloud model”

37. parcel mixing ratio=5g/kg Expansion, (adiabatic process) gives condensation temperature Cooling: (Isobaric process) gives dew point temperature

38. Wet Bulb Temperature

39. Equivalent Potential temperature conserved in adiabatic motions Parcel at 850 hPa, T=12.5oCr=6 g/kg Te qe (=315K) Raise parcel pseudoadiabatically until all humidity condenses and then descend dry adiabatically to reference pressure

40. Dry adiabatic processes Moist adiabatic processes potential temperature equivalent potential temperature dry static energy moist static energy water vapour mixing ratio total water mixing ratio Summary: Conserved Variables(approximately, r.e. Emanuel 1994)

41. Practice! • 1. Plot an air parcel at 750 hPa which has • T=10C • q=2 g kg-1 • State the dew-point temperature • 2. If it is subjected to forced ascent, • State the pressure at which it saturates • 3. The parcel continues to rise pseudoadiabatically to 200hPa, what is its temperature and humidity (r) there? • 4. The parcel then descends back to 750hPa, what is the final temperature and r?

42. Practice! Results 1 - 10 of about 6,060,000 for practice makes perfect [definition]. (0.05 seconds) 1. The environment at 750 hPa has T=10C and r=2 g kg-1. A saturated thermal arrives at this level with the temperature T=5C and entrains the environmental air with equal ratios (1-1), what is the T/r of the mixed parcel? 2. Then precipitation falling through the layer evaporates and brings this new parcel to saturation, what is the temperature?

43. Finally: A few Unofficial Social Tips…see also www.reading-guide.co.uk For Music listings: pick up “Bla Bla” Jonglers, Salsa Sunday Evenings, Lessons all levels (19.30? £?) free dancing after Indian Restaurant Jazz club – free live music on Thurs “Thai corner” – family run Thai food “3Bs” (basement under town hall) – Free music Thurs, great guest beers Station If the Weather is nice, don’t forget to take a walk by Thames, off the top of this map, or take a train to Pangbourne or Goring nearby and see the Thames there ARTS cinema at Reading University, Tues/Thurs, see readingfilmtheatre.co.uk For London theatre, check out OFFICIAL half-price ticket booth in Leicester Square “Beijing noodle house” – Chinese food, family run Abbey Ruins, Reading’s (only) historical part! Iguana: Coffee+cocktails (upstairs) HobGoblin, smoky - but character pub “Sweenie and Todds” Pie Pub After Dark, grungy nightclub Wagamamas (Oracle by canal), Asian noodle Chain, not bad though “RISK”: Salsa 19.30 lessons £5/9, free dancing after 21:30 Tuesday (upstairs) International drinks

44. Answer Sheet for Lesson 1 • Sheet 1 • Td = -10oC • 550 hPa • T= -78oC • For the humidity it is not possible to read from this tephigram. Therefore you have to use the Clausius Clapeyron equation and the definition of mixing ratio to find q=6x10-6 kg kg-1 using T=-78oC and p=200 hPa • T=15oC, so the parcel has gained 5oC from the latent heating released by the condensation process. • Since humidity is conserved then q =6x10-6 kg kg-1 • Sheet 2 • T=12.5oC and q=4.7 g kg-1. Note that the humidity scale is not linear! • T=4oC