Synotic Meteoroloy I Some hermodynamic Concets
Geootential Heiht Geootential Heiht (h): the otential enery of a nit mass lifted from srface to. Φ d 0 -Since constant in the trooshere, we can write Φ Δ m m m 2 2 s s 2 J k We can now define eootential heiht!
Geootential Heiht h 1 Φ d.8 9.8 9 0 (no nits) eootential meters (m) m 2 /s 2 m 2 s -2 denotes a otential enery Ultimately, h is emloyed to ensre NO v in horiontal flow.
Geootential Heiht Here s how it works: i. Earth is an oblate sheroid R R e R e > R
Geootential Heiht ii. As sch, increases as yo travel S -> N (yo re decreasin distance from the center of mass) iii. Frther, srfaces at a constant heiht above the earth (sea level) will also be misshaen. 10,000m 000m MS * So, ravity chanes, y on those srfaces!
Geootential Heiht lower over ole eootential srface srface A v hiher over eqator v On a srface, is not always a (arallel allel to local kˆ ) If we assme it is, then we se a v that does not oint to the center of Earth s mass. Doin so, artificially introdces an eqatorward acceleration, A v (Moral: Use h).
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 tends to the eqator) On a eootential srface N 0 o Φ constant
Geootential Heiht On a eootential srface N 0oΦ constant ball is at rest eootential srface NO comonent of v actin alone the srface v does not affect horiontal flow On constant srfaces, we se lines of constant h (m), e.., 5640 m at 500 mb. A constant srface does not eqal a eootential srface, BU v is to each heiht line.
Geootential Heiht 982 9.82 9970m 9.83 9.79 9.80 977 9.77 10,021m 10,000m 9.78 o 9.84 o 9.81 90 o Δ N h 9.8 45 o m 9970m 9.83 2 s 9.8 10,000m m 10,000m 45 o Δ h 9.80 10,000m 98 9.8 s 98 9.8 2 m 10,021m 0o Δ h 9.78 10,000m 2 9.8 s 9.8 0 o o 9.79
Geostrohic Winds y f ρ 1 x f v ρ 1 - coordinates f Φ 1 v y f y f Φ 1 - coordinates x f x f v
Geostrohic Winds 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 Δh more accrately from RAOBS. Δh R 9.8 v ln U eliminated variable, Δ R v ln U
Hysometric Eqation Derivin the hysometric eqation or thickness eqation assme a hydrostatic atmoshere ρ and introdce the eqation of state: ρ 1 R 2 Rd
Hysometric Eqation Now, sbstitte 2 into 1 R d R l d 1 Now, assme d R l d 1 d R l d d R ln
Hysometric Eqation [ ] ) ( ln ln R R d ) ( ln ln R R d ) ( ln R d R ) ( ln d R ) ( ln and that brins s to...
Hysometric Eqation R d Δ ln he Hysometric Eqation! Ex 1: 287 Δ 9603.7 J KK 9.81 km ks m 2 s 2 2 (269 K ) ln m 2 s 9603.7 m 1016 300 mb mb msl 297 (75 o F) 300 242 (- 24 o F) 269.5
Hysometric Eqation Ex. 2: 287 Δ Δ J KK R d ln (285 K ) 1000 ln m 500 9.81 2 s mb mb 1000 303 K 500 267 K 500 285 K 5779 km ks m s 2 2 2 5779 m known 500mb (1000mb) known
Hysometric Eqation Ex. 3: Δ R d Δ ln Δ R d R d e Δ 450 m,, ln e ( 450 ) ( 303 ) R Δ 9.8 287 954 ha e 1003. 6 ha d 954 ha known Δknown MS
Hysometric Eqation So Hysometric Eqation is Vital for: 1. Redcin Station ressre to sea level. 2. Determinin heihts of er-level ressre srfaces.
Hysometric Eqation *Crcial Qestion* Δ R d constant constant R d ln If we fix and, then Δ is deendent only on. Δ α
Hysometric Eqation 5220 5280 5340 5400 cold 5160 thickness attern warm thickness stron radient of stron radient of r lare Δ
hickness and hermal Wind hickness and thermal wind: 1. hels exlain how winds chane in vertical 2. assmes eostrohic atmoshere tells s somethin abot vertical strctre of 3. tells s somethin abot vertical strctre of atmoshere
hickness and hermal Wind v v v V V - V V r V r V shear of V r eostrohic cold wind -3-2 -1 V r 1 kˆ V v R f d ln kˆ n- 3 n- 2 n-1 n warm thickness thermal wind blows ll to thickness contors
hickness and hermal Wind chanes in the wind in the vertical (vertical wind shear) α the mean thermal radient over the layer V r r V r V V r V r V r V V r
hickness and hermal Wind With mltile ressre levels: backin conterclockwise (cold air always to left) V v V300 cold V r V r 850 warm * backin with heiht mean layer V is casin cold advection
hickness and hermal Wind Veerin- clockwise chane of wind warm V v 850 r V cold V r 300 * veerin with heiht mean layer V is casin warm advection
otential emeratre Consider the First aw of hermo dh C d- α d 1 First aw states that, for a closed system, no heat may be added to/removed from a system (arcel). dh 0 C d α d 2 Now, eqation of state ives s ρr or ρ /R or α R/ 3
otential emeratre Sbstitte 3 into 2 to et C d R d Next, divide by to et C d R d d Now divide throh by C to et Rd 287.06 J/k K C 1005 J/k K d Rd C d
otential emeratre We then et θ d Rd C 1000 d d θ tem at 1000mb Τ tem at some ressre level ln θ -ln Rd ln 1000 ln C ln θ Rd C ln 1000
otential emeratre ake exonent of both sides θ Rd C 1000 let θ 1000 Rd C θ κ 1000 θ otential tem. tem. a arcel will have if broht down to 1000mb from some level, havin some tem.,.
otential emeratre Examle θ 1000 κ Given, θ 28 o C301K, what will be at 850 mb? 301K 1000 850.286 301K 1.04758 287.3 K 14.3oC