Synoptic Meterorology I. Some Thermodynamic Concepts

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1 Synotic Meteroroloy I Some hermoynamic Concets

2 Geootential Heiht Geootential Heiht (h): the otential enery of a nit mass lifte from srface to. Φ 0 -Since constant in the trooshere, we can write Φ m m m 2 2 s s 2 J k We can now efine eootential heiht!

3 Geootential Heiht h Φ 9.8 (no nits) eootential meters (m) m 2 /s 2 m 2 s -2 enotes a otential enery Ultimately, h is emloye to ensre NO v in horiontal flow.

4 Geootential Heiht Here s how it works: i. Earth is an oblate sheroi R R e R e > R

5 Geootential Heiht ii. As sch, increases as yo travel S -> N (yo re ecreasin istance from the center of mass) iii. Frther, srfaces at a constant heiht above the earth (sea level) will also be misshaen. 10,000m MS * So, ravity chanes on those srfaces!

6 Geootential Heiht lower over ole eootential srface srface A v hiher over eqator On a srface, v is not always (arallel to local kˆ ) If we assme it is, then we se a v that oes not oint to the center of Earth s mass. Doin so, artificially introces an eqatorwar acceleration, A v (Moral: Use h).

7 Geootential Heiht On a constant heiht srface (above MS) N 0 o 10,000m ole > eqator * is not always to (a ball on this srface tens to the eqator) On a eootential srface N 0 o Φ constant

8 Geootential Heiht On a eootential srface N 0oΦ constant ball is at rest eootential srface NO comonent of v actin alone the srface v oes not affect horiontal flow On constant srfaces, we se lines of constant h (m), e.., 5640 m at 500 mb. A constant srface oes not eqal a eootential srface, BU v is to each heiht line.

9 Geootential Heiht eootential sfc ,021m 10,000m 9970m o o 45 o m 9970m N h s 9.8 m 45o h s m 0o h s 10,000m ,021m 9.8 o ,000m 10,000m 10,000m 0 o o 9.79

10 Geostrohic Wins x f v y f ρ ρ 1 1 x f x f v y f y f Φ Φ coorinates - coorinates

11 Geostrohic Wins Qick Conversion Rle (or h) 1. relace 1/ρ with 2. exchane with (eometric heiht) or h (eootential heiht) *We se eootential heihts to et h more accrately from RAOBS. h R 9.8 v U eliminate variable, R v U

12 Hysometric Eqation Derivin the hysometric eqation or thickness eqation assme a hyrostatic atmoshere ρ an introce the eqation of state: 1 ρ R 2

13 Hysometric Eqation Now, sbstitte 2 into 1 R ρ R l R l 1 R l Now, assme R

14 Hysometric Eqation [ ] ) ( R ) ( R ) ( R ) ( R an that brins s to...

15 Hysometric Eqation R he Hysometric Eqation! Ex 1: 287 J KK 9.81 (269 m 2 s K ) mb mb msl 297 (75 o F) (-24 o F) km ks m 2 s m 269.5

16 Hysometric Eqation Ex. 2: R 287 J KK 9.81 (285 m s 2 K ) mb mb K K 285 K km 2 500mb 5779 ks m m known s 2 (1000mb) known

17 Hysometric Eqation R ( ) ( ) ha e ha ha m e e R R R ,, MS known known Ex. 3:

18 Hysometric Eqation So Hysometric Eqation is Vital for: 1. Recin Station ressre to sea level. 2. Determinin heihts of er-level ressre srfaces.

19 Hysometric Eqation *Crcial Qestion* R constant constant R If we fix an, then is eenent only on. α

20 Hysometric Eqation col 5160 thickness attern warm thickness stron raient of stron raient of r lare

21 hickness an hermal Win hickness an thermal win: 1. hels exlain how wins chare in vertical 2. assmes eostrohic atmoshere 3. tells s somethin abot vertical strctre of atmoshere

22 hickness an hermal Win v v v V V - V shear of V r r V r V eostrohic col kˆ V r win V v R f kˆ n- 3 n- 2 n-1 n warm thickness thermal win blows ll to thickness contors

23 hickness an hermal Win chanes in the win in the vertical (vertical win shear) α the mean thermal raient over the layer V r V r V r V r V r V r V r

24 hickness an hermal Win With mltile ressre levels: backin conterclockwise (col air always to left) v V300 col V r V r 850 warm * backin with heiht mean layer V is casin col avection

25 hickness an hermal Win Veerin- clockwise chane of win warm v V850 V r col V r 300 * veerin with heiht mean layer V is casin warm avection

26 otential emeratre Consier the First aw of hermo h C - α 1 First aw states that, for a close system, no heat may be ae to/remove from a system (arcel). h 0 C α 2 Now, eqation of state ives s ρr or ρ /R or α R/ 3

27 otential emeratre Sbstitte 3 into 2 to et C R Next, ivie by to et C R Now ivie throh by C to et R J/k K C 1005 J/k K R C

28 otential emeratre We then et θ R C 1000 θ tem at 1000mb Τ tem at some ressre level θ - R 1000 C θ R C 1000

29 otential emeratre ake exonent of both sies θ R C 1000 let θ 1000 R C θ 1000 K θ otential tem. tem. a arcel will have if broht own to 1000mb from some level, havin some tem.,.

30 otential emeratre Examle θ 1000 K Given, θ 28 o C301K, what will be at 850 mb? 301K K K 14.3oC

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