StableMixing ratio35 Part 3. Atmospheric Thermodynamics The Gas Laws Eq. of state pV mRT= mass gas constant for 1 kg of a gas /1/mvpRTorpRTwhereρρααρ==== For dry air *,dadddpRTwhereRRMρ==universal gas constant (8314.3 1degJkmol−)36 .weighted over molecular wt./iidiiimMmM==∑∑ iM molecular weight= 11287 degdRJkg−−= For water vapor: vvveRTρ= *11461 degvwRRJkgM−−== Define: 0.622dwvdRMRMε== = Dalton’s Law: idpppe=∑ = + ''dvdvmmVρρρ+==+37 ''''11(1)vv d dddvdvdvddddeRTandpRTpe eRT RTpe eTRTRRRpeeRT p ppeRT pρρρρ ρερε==−=+= +−=+=−+=−− Define /1 (1 )veTTpε=−− Then dvpRTρ= Virtual temp: T that dry air must have in order to have the same density as moist air at the same pressure.38 Adding moisture to air has the effect of raising Tv Moist air is less dense than dry air. Hydrostatic Equation Hydrostatic pressure is a result of the weight of an air column. Consider an air slab having unit cross-sectional area:39 Weight of slab: mg g dzρ= p is a force/area. Therefore, δp represents a net force acting in the vertical direction, and since p decreases with height, -δp acts upward. To be in equilibrium: pg zδρδ−= or dpgdzρ=− hydrostatic equation40 Alternate view Vertical equation of motion: 1Viscositydw dpgdt dzρ=− − + acceleration P-grad gravitational in vertical forceacceleration acceleration Ignoring viscosity, assume 0dwdt= Then 1ordpgdzdpgdzρρ−==−41 To find p at any height z, (0) 0()()ppz zzdp gpdzorpz g dzρ=∞∞−==∫∫∫ Geopotential φ - is the work that must be done against gravitational field to raise a mass of 1 kg from sea level to a given height.42 dgdzφ= (Change in potential energy) from the hydrostatic relation ordp g dz gdz dpddpραφα−= → =−=− The geopotential at height z is ()(0) 0 0()zzzdgdzφφφφ===∫∫ We define geopotential height 000() 1zzz gdzggφ==∫43 where g0 is globally-averaged g at earth’s surface 209.8gms−= Z is what is plotted on our weather maps. Express Z in terms of T,p d dp gdzφα=− = Eq. Of state: dorR=dvvpRTTpρα= Then lndvdRTdpφ=− Integrating from 2112 1 221withlnpdvpppRTdpφφφφ→→−=−∫44 Dividing by g0, we have 21210pdvpRZZ Tdlnpg−=−∫ Replace Tv by average vTthrough a layer, 21 210(/)orvdRTZZ lnppg−=− 21 120(/)vdRTZZ lnppg−= Called thickeness eq.45 The geopotential height of the 500 mb surface is sea level500500lnvdoRT pZgp Z500 will be low if psea level is low. Conversely, for a given psea level, Z500 is low, if mean Tv between surface and 500 mb is low.46 First Law of Thermodynamics Internal Energy = Sum of kinetic energy of molecules. Increases in internal energy in the form of molecular motions is manifested as increases in temperature. Consider a unit mass of gas which absorbs a certain quantity of heat energy q (in joules) – i.e., by radiation or thermal conduction. As a result, the gas may do a certain amount of external work W. The excess of energy supplied to the gas over the external work done by the gas is 21.qw uu u−=∆= − In differential form: dq dw du−= (First Law of Thermodynamics) incremental differential differential heat elemental internal work energy Consider a gas contained in a cylinder of fixed cross-sectional area with a moveable, frictionless piston. Vd∝ since area is constant47 The work done by the gas in expanding is equal to the force exerted on the piston (pA) multiplied by the distance dx through which the piston moves. Thus, }dVdw p Adx pdV== called p-V work For a unit mass of gas in the free atmosphere we have dw pdα= or }dwdq du pdα=+ First law48 Specific Heats Suppose that a small amount of heat dq is given to a unit mas of gas and its temperature increases from T to T + δT. The ratio dq/dT is called the specific heat. If the volume is held fixed then .vconstdqCdTα== If α=const. 0 dq du pdα=+49 .vconstduCdTα== but for an ideal gas, ()onlyuuT= then vduCdt= or vdq C dT pdα=+ specific heat at const. p .()since ()ppconstvvvdqCdTdq C dT pdCdT d p dppRTdq C dT d RT dpααααα===+=+ −==+ −50 or ()if .()orvvvpppvdq C R dT dpp constdqdq C R dT C R CdTCCRα=+ −==+ ⇒ =+==+ Thus p=Cdq dT dpα− Enthalpy Define: as enthalpy()hupdh du d pαα=+=+51 Remember vdu C dT= From 1st Law: ()(Rem: )vdhdq C dT pddu d p dpdq dh dpdq cpdT dpααααα=+=+ −=−=−14243 Thus pdh C dT= or phCT= with h=0 @ T=052 Previously we defined geopotential ()( )pd gdz dpdq dh dpdh dCTφααφφ==−∴=−=+= + Called Montgomery stream function If no heat is added or taken away from a parcel in a hydrostatic atmosphere, then 0.dq h constφ=⇒+= The Montgomery stream function is conserved along isentropic surfaces.53 Potential Temperature It is often convenient to define a variable that is conservative under adiabatic motion. Consider an adiabatic process: 0pdq C dT dpα== − Using the eq. of state, pRTα= Then, 0nnppCdpCdT RT d T d ppR−== −ll integrating from p00=1000 mb where we let T=θ, to p, 00pTpTpCdnT dnpRθ==∫∫ll 00/00ppCRCnT n p pRorTppθθ==ll /00(/)pRCTp pθ= Poisson’s Eq. The potential temperature is the temp. that the parcel of air would have if it were compressed adiabatically from its initial level (p,T) to sea level pressure 001000pmb=.54 The Adiabatic Lapse Rate Consider an infinitely small parcel of air that is: • Thermally isolated from its environment such that heat is not added or taken away from the parcel (adiabatic). • The parcel immediately adjusts to the hydrostatic pressure at any level. • Its motion is small so that its K.E. remains small.55 For a parcel experiencing adiabatic transformations, 0pdq C dT dpα== − Take derivative with respect to Z, 0pdT dpCdz dzα−= Since pressure adjusts to hydrostatic p at any level, the hydrostatic equation. Gives us: or0/ppdpgdzdT dTCg gCdz dzρ=−+=⇒ =− Define: /9.8/dpgC Ckmγ==o56 Parcel Stability The buoyancy of a parcel relative to its environment is defined as: '00'vTBoy g gTθθ(( Vertical accelerations are then, ''00dw