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MET 60

MET 60. Chapter 3: Atmospheric Thermodynamics. In chapters 3, 4, and 6, we will be zooming in and out from the largest-scale to the molecular/atomic scale (ditto chapter 5 but…) http://www.youtube.com/watch?v=Sfpb9GqYLiI Ditto chapter 5 but …. The layout of chapter 3 is:

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MET 60

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  1. MET 60 Chapter 3: Atmospheric Thermodynamics MET 60 topic 03

  2. In chapters 3, 4, and 6, we will be zooming in and out from the largest-scale to the molecular/atomic scale (ditto chapter 5 but…) http://www.youtube.com/watch?v=Sfpb9GqYLiI Ditto chapter 5 but … MET 60 topic 03

  3. The layout of chapter 3 is: Some thermodynamics (“thermo” or “TD”) • Application(s) More thermo • More applications Yet more thermo • Yet more applications Etc. MET 60 topic 03

  4. Basic Thermo: The Ideal Gas Law (or the Equation of State) p = pressure;  = density; Rd = gas constant (???); T = temperature. “Used” in virtually everything – i.e., it is almost always assumed! • Application: The Hydrostatic Equation MET 60 topic 03

  5. Also – geopotential geopotential height thickness • Thermo: The First Law of Thermodynamics Also: specific heat enthalpy (?) MET 60 topic 03

  6. Applications: adiabatic processes dry adiabatic lapse rate potential temperature thermodynamic diagrams • Water Vapor in the air Measures of vapor in the air mixing ratio specific humidity relative humidity dew point MET 60 topic 03

  7. Water Vapor in the air (cont.) lifting condensation level (LCL) latent heat Chinook winds – moist ascent followed by dry descent • Application: Static stability As a parcel of air is lifted (or rises), what can happen to it? MET 60 topic 03

  8. If T(parcel) > T(environment), parcel rises (A) If T(parcel) < T(environment), parcel sinks (B) (A) Is termed an unstable situation, and air parcels can rise  deep clouds (e.g., Cb) (B) Is termed stable situation, no deep clouds BUT gravity waves in cloud patterns Fig. 3.14 MET 60 topic 03

  9. Thermo: The Second Law of Thermo entropy Clausius-Clapeyron Equation MET 60 topic 03

  10. The Ideal Gas Law (or the Equation of State) It is experimentally found that all gases obey this relation: (3.1) p = pressure (Pa – not mb or hPa) V = volume (m3) m = mass (kg) T = temperature (degrees K – not C or F) R is a constant of proportionality (the gas constant) MET 60 topic 03

  11. Value of R depends on the gas involved – see below. We also write: where  = density (kg/m3), or: where  is specific volume = 1/. MET 60 topic 03

  12. Two special cases (which we often look at!): • Constant temperature… isothermal … Volume is inversely proportional to _____________ This is Boyle’s Law (1600’s !!) • Constant pressure… Volume is proportional to _____________ This is Charles’ Law (late 1700’s – early 1800’s) MET 60 topic 03

  13. Constant volume… Pressure is proportional to _____________ This is also Charles’ Law MET 60 topic 03

  14. For dry air, where Rd = gas constant for dry air. Value/meaning of Rd? MET 60 topic 03

  15. Consider any (pure) gas… a) It has a molecular weight = M b) A mole (or mol) of this substance is defined as the number value of M, expressed in grams Examples: oxygen has M = 32 – one mol of O2 weighs 32 grams water has M = 18.015 – one mol of water weighs 18.015 grams CO2 has M = 44.01 – one mol of CO2 weighs 44.01 grams MET 60 topic 03

  16. For all substances, a mol of the substance has the same number of molecules – NA (Avogadro’s number) Thus in Eq. (3.1), for 1 mol of gas, the constant (R) is the same for all gases. This is the Universal Gas Constant (R*). So for 1 mol: pV = R*T (unit mass) And for n moles: pV = nR*T (mass “n”) MET 60 topic 03

  17. Now if we have a mass m of the gas, then: m = n M Thus: So that (R*/M) serves as a “gas constant for that species”. So – for dry air – we just need the value of M – call it Md! MET 60 topic 03

  18. We know the components of dry air (78% N2, 21% O2 etc.) and we know the molecular weight of each of these. We compute Md from (3.10) to get Md = 28.97 And thus, Rd = R*/ Md = 287 J K-1 kg-1 MET 60 topic 03

  19. Recap… Ideal Gas Law / Eqn. Of State for dry air: Rd = R*/ Md = 287 J K-1 kg-1 For moist air??? Look at pure water first – then combine. MET 60 topic 03

  20. For water vapor… Mw = 18.016 So Rv = R*/Mw = 461.5 J K-1 kg-1 And the ideal gas equation is ev = RvT (3.12) where e = pressure due to water vapor alone v is the specific volume of the vapor and Rv is the gas constant for water vapor MET 60 topic 03

  21. For moist air? We use Dalton’s Law of partial pressures: Total pressure = sum of individual pressures (as long as the gases do not interact chemically!) Example 3.1 MET 60 topic 03

  22. For moist air, we could use an Rv but the value would depend on how much moisture is in the air (not constant!) So Rv would be a “variable constant”!!! Instead, we introduce: Virtual Temperature (Tv) A fictitious temperature dry air would need to have in order to have the same density as moist air @ same pressure MET 60 topic 03

  23. Note:  (moist air ) <  (dry air) Because… Also… Tv > T (for moist air) In practice, T and Tvdiffer by only a few degrees MET 60 topic 03

  24. W&H derive:  = Rd/Rv = 0.622 MET 60 topic 03

  25. The Hydrostatic Equation Consider a parcel of (dry) air… p, , T pe, e, Te MET 60 topic 03

  26. Suppose p > pd p, , T pe, e, Te MET 60 topic 03

  27. The parcel expands! and vice versa if p < pd p, , T pe, e, Te MET 60 topic 03

  28. What mattered was the pressure difference or pressure gradient -not the actual pressure Now apply this thinking to a layer of air -Fig. 3.1 MET 60 topic 03

  29. The hydrostatic equation (remember it!) And of course at the same time: MET 60 topic 03

  30. Some applications… using the hydrostatic equation and the equation of state Look at a 500 mb/hPa map MET 60 topic 03

  31. MET 60 topic 03

  32. The contours link locations where the 500 hPa surface is at the same “height” ASL. Expressed in decameters (dm or dam) Example: a height of 5000 m (5 km) is contoured as 500 dm As we will see, height(500 hPa)  temperature of (1000-500 hPa) layer thickness MET 60 topic 03

  33. Geopotential From the 500 mb chart, consider a parcel on the 576 dm contour. To get there means that work has been done against the force of gravity. Geopotential is defined as the work done to raise a mass of 1 kg from altitude z=0 to a desired altitude (z). MET 60 topic 03

  34. z+dz Geopotential done = force x distance work = (mass x acceleration) x distance = (1 x g) x dz = gdz We define geopotential, , by: d = gdz z MET 60 topic 03

  35. d = gdz And if we treat g as constant and add up over z (“integrate”), we get: Units? m2s-2 – energy units To make these units and values more meaningful, we define:  = gz MET 60 topic 03

  36. Geopotential height: Z =  /9.8 = {gz / 9.8} m2s-2 For the troposphere, Z  z. We even define a geopotential meter (gpm) such that: at an altitude z meters, the geopotential height is Z gpm, where the numerical values are about the same Example: at height 5000 meters, Z  5000 gpm height units energy units MET 60 topic 03

  37. Z=5822 gpm MET 60 topic 03 and z5822 m !!

  38. z2 Thickness of a layer We can also talk about thickness… Thickness = Z2 - Z1 z1 MET 60 topic 03

  39. Now we recall: So: So by integration: (thickness) And we need values of Tv to get any further… MET 60 topic 03

  40. Suppose Tv is constant ( ) in a layer (isothermal). Then: Two things: • Thickness  (layer mean) temperature – obvious! • If we set: MET 60 topic 03

  41. Then: Rearranging: This is basically the same as Eq.(1.8)!!! As before, H is scale height, and now MET 60 topic 03

  42. What if Tv is NOT constant (which it will not be)? We can still define a layer-average (Eq. (3.28)) and write: where go is a constant value of g (9.8). This is called the hypsometric equation. -- Use above form to find heights given pressures. -- Invert and use to find pressures given heights. MET 60 topic 03

  43. Warm-core lows… Warm at its center Example: a hurricane (Fig. (3.3)) Intensity decreases with height Cold-core lows… Cold at its center Example: a mid-latitude cyclone Intensity increases with height MET 60 topic 03

  44. The First Law of Thermodynamics Consider a closed system (e.g., a parcel of air). It has internal energy (“u”) = energy due to molecular kinetic and potential energies Suppose some energy (dq) is added to the system Example: via radiation from the sun What happens? MET 60 topic 03

  45. Some of the energy goes into work done (dw) by the system against its surroundings Example: expansion What’s left is a change in internal energy: This defines du and we will show that du  T du = dq – dw MET 60 topic 03

  46. -add dq of energy -parcel may expand (dw) -internal temperature may change (du, dT) 1) add dq 3) possible T change internally 2) possible expansion MET 60 topic 03

  47. du = dq – dw So… Write more usefully… For us, the main work (dw) is expansion/contraction work: Use: • work = force x distance • Pressure = force per unit area • Assume unit mass where mass = density x volume To show that: dw = pd MET 60 topic 03

  48. dq = du + pd Thus… which is closer to useful (see more below…) Next useful concept…specific heat Suppose we add some thermal energy (dq) to a unit mass of a substance Water Air Soil MET 60 topic 03

  49. We expect T(substance) to increase How much? We can define specific heat as: More carefully: constant volume constant pressure heat added temp change MET 60 topic 03

  50. Specific heat is the heat energy needed to raise the temperature of a unit mass of substance by one degree. Values (p. 467!!)? MET 60 topic 03

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