1. Static Stability. (ρ V ) d2 z (1) d 2 z. = g (2) = g (3) T T = g T (4)

Size: px

Start display at page:

Download "1. Static Stability. (ρ V ) d2 z (1) d 2 z. = g (2) = g (3) T T = g T (4)"

Laurence McDowell
5 years ago
Views:

1 1. Sttic Stbility 1. Sttic Stbility of Unsturted Air If n ir prcel of volume V nd density ρ is displced from its initil position (z, p) where there is no net force on it, to (z + z, p p). From Newton s Second Lw of Motion F = m we hve (ρ V ) d2 z dt (ρv ) g(ρ V ) 2 }{{}}{{} Boyncy Force Weight ( ) ρ ρ dt 2 ρ ( ) 1/T 1/T 1/T ( ) T T T (1) (2) (3) (4) If T 0 is the environmentl temperture t z, the new temperture of the ir prcel t z+z will be T = T 0 Γ d z, nd the mbient temperture will be T = T 0 Γz, so T T = (Γ d Γ)z. So the eqution of motion cn be written: dt 2 = g T (Γ d Γ)z (5) dt 2 + N 2 z = 0 (6) Where N = [ g T (Γ d Γ) ] 1/2 is the Brunt-Visl frequency. Eqution 6 is second order ordinry differentil eqution. Remember: A second order ordinry differentil eqution of the form y + by + c = 0 hs solution: The three possible cses re: 1

2 Roots of r 2 + br + c = 0 r 1 nd r 2 rel nd distinct r 1 = r 2 = r r 1,r 2 complex α ± iβ Generl solution y = c 1 e r1x + c 2 e r 2x y = c 1 e rx + c 2 xe rx c 1 e (α+iβ)x + c 2 e (α iβ)x N 2 > 0 if Γ < Γ d, dθ/dz > 0 N 2 < 0 if Γ > Γ d, dθ/dz < 0 N 2 = 0 if Γ = Γ d, dθ/dz = 0 i. Stble (subdibtic) N 2 > 0 In this cse, the solution to Eqution 6 hs solution of the type: z (t) = c 1 e int + c 2 e int (7) c 1 nd c 2 re determined by the initil conditions. The solution is the Brunt- Visl oscilltion t the frequency N nd period τ b = 2π/N ii. Neutrl Stbility N 2 = 0 In this cse, the solution to Eqution 6 hs solution of the type: ( ) dz z (t) = z (t = 0) + t (8) dt t=0 (using the initil conditions). Consequently, becuse the density of the ir prcel nd mbient tmosphere re the sme, the buoyncy force nd wight re equl nd opposite nd there is not net force on the ir prcel. iii. Unstble (subdibtic) N 2 < 0 In this cse, the solution to Eqution 6 hs solution of the type: z (t) = c 1 e N t + c 2 e N t (9) c 1 nd c 2 re determined by the initil conditions. The prcel continues moving in the direction in which it ws displced. When n environmentl lyer is unstble, dry convection occurs with wrm ir prcels rising from the bottom of the lyer nd cold ir prcels sinking from the top. 2

3 Tble 1: Summry for Unsturted Air Equilibrium Lpse Rte dθ/dz Stble Γ < Γ d > 0 Neutrl Γ = Γ d = 0 Unstble Γ > Γ d < 0 A lyer within the troposphere in which Γ < 0 or dt/dz > 0 is clled n inversion lyer where the ir is stbly strtified. If the tmospheric boundry lyer is cpped by n inversion lyer, the ir within the boundry lyer cnnot penetrte verticlly cross the inversion lyer. 1b. Stbility of Sturted Air As sturted ir prcel is displced upwrds, its temperture will decrese t the pseudodibtic lpse rte. If the mbient lpse rte is greter thn the pseudodibtic lpse rte, Γ > Γ s, the displced prcel will be wrmer thn its environment nd will be ccelerted in the direction of the displcement. This ir is unstble with respect to the pseudodibtic prce ldisplcement. If we llow for condenstion on scent, five regimes cn exist: Tble 2: Summry for Unsturted Air Equilibrium Absolutely Stble Neutrl (sturted) Conditionlly Unstble Neutrl (unsturted) Absolutely Unstble Lpse Rte Γ < Γ s Γ = Γ s Γ d > Γ > Γ s Γ = Γ d Γ > Γ d 1c. Ltent Instbility Let s consider conditionlly unstble lyer. When n ir prcel rises verticlly, the work per unit mss performed is: 3

4 with d2 z dt 2 w = 1 m ( ) T T T. F ds = 1 m F dz = dz (10) dt2 T T dz w T dp dp = R for A = T d( ln p) nd A = T d( ln p). (T T )d( ln p) (11) w = R(A A) (12) Tble 3: Emgrm (Energy per unit mss digrm) A digrm of T vs ln p is true thermodinmic digrm on which the re is proportionl to energy (or work). The re enclosed in ny contour is proportionl to the work done in cyclic process defined by the contour. dw = pdα = RdT αdp = RdT RT d ln p dw = RdT RT d ln p = RT d ln p The most elementry pproch to find the verticl velocity of convective element is: The work per unit mss will be trnsformed into kinetic energy: w = so d 2 z b dt dz = dv dz 2 dt dt dt = v dv dt dt = 1 2 dv 2 = 1 2 (v2 b v 2 ) (13) 1 2 (v2 b v 2 ) = R(A A) (14) When the process curve C lies to the right of the sounding curve C, the buoyncy force exceeds the weight of the ir prcel, hence A > A, w > 0 nd v 2 b > v2. In this cse, there is upwrd ccelertion. If the process curve lies to the left of the sounding curve C, work equl to the negtive re on the emgrm must be done on the ir prcel to cuse it to scent. 4

5 Figure 1: Figure on pge 63 of notes. Figure on pge 61 of notes. In order for the ir prcel to follow pth C 1, we must perform A work on the prcel (orogrphic scent, convergence). The level t which the ir prcel reches sturtion, point P s is the Lifting Condenstion Level (LCL). When the ir prcel surpsses the level t which the process curve C crosses the sounding t the Level of Free Convection (LFC), the net buoyncy force performs work proportionl to re A+. Convection will continue freely until C 1 crosses C gin, nd the prcel decelertes. The totl potentil energy vilble for conversion to kinetic energy is termed the convective vilble potentil energy (CAPE) pc CAP E = R (T T )d( ln p) p LF C (15) The instbility is lrger for lrger T nd r. r increses : Segment P 1 P s becomes shorter, A decreses nd A+ increses. Ground T increses : The lpse rte becomes steeper, the lyer becomes unstble nd verticl mixing strts. Mixing results in dry dibtic lpse rte for the lyer. The lowest prt of the sounding becomes substituted by dry dibts, nd the top of the lyer cn rech sturtion, this is clled the Convective Condenstion Level (CCL). From tht moment on, convection my proceed spontneously long the dry dibt without ny need for forced lifting. We find the bses of cumulus clouds t the CCL In prctice, the verticl velocity nd penetrtion height of cumulus tower re reduced from dibtic vlues by nonconservtive effects. Cooler nd drier environmentl ir tht is entrined into nd mixed with moist therml depletes scending prcels of positive buoyncy nd kinetic energy. 5

1. Static Stability. (ρ V ) d2 z (1) d 2 z. = g (2) = g (3) T T = g T (4)

1. Static Stability. (ρ V ) d2 z (1) d 2 z. = g (2) = g (3) T T = g T (4) NCAR (National Center for Atmospheric Research) has an excellent resource for education called COMET-MetEd. There you can find some really great tutorials on SkewT-LogP plots: visit http://www.meted.ucar.edu/mesoprim/skewt/index.htm.