A Discre)za)on of Deep Atmospheric Model Dynamics for NCEP Global Forecast System

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1 A Discre)za)on of Deep Atmospheric Model Dynamics for NCEP Global Forecast System Hann- Ming Henry Juang Environment Modeling Center, NOAA/NWS/NCEP, Washington, DC 10 April 2014 PDEs 2014, H. Juang 1

2 Introduction While NCEP GFS extends its vertical domain to couple with space environmental model, we called this version of GFS as Whole Atmospheric Model (WAM). WAM is a hydrostatic system with enthalpy as thermodynamic variable (Juang 2011 MWR). We propose to do deep atmospheric dynamics for NCEP GFS to support WAM Here, we would lie to re-iterate the reasons to use deep atmospheric dynamics and to illustrate the discretization of deep atmospheric dynamics for NCEP GFS. 10 April 2014 PDEs 2014, H. Juang 2

3 Opr GFS vs WAM 64 layers 150 layers 10 April 2014 PDEs 2014, H. Juang 3

4 Example of T profile of 150 layers WAM uses generalized hybrid coordinate with enthalpy CpT as thermodynamics variables, where Cp is summaton of each gases. R Cp O O O Dry air H2O April 2014 PDEs 2014, H. Juang 4

5 Maxima wind (m/s) at NCEP GFS 150 layers WAM in do_dynamics_two_loop for spdmx at = spdmx(001:010)= spdmx(011:020)= spdmx(021:030)= spdmx(031:040)= spdmx(041:050)= spdmx(051:060)= spdmx(061:070)= spdmx(071:080)= spdmx(081:090)= spdmx(091:100)= spdmx(101:110)= spdmx(111:120)= spdmx(121:130)= spdmx(131:140)= spdmx(141:150)= April 2014 PDEs 2014, H. Juang 5

6 Shallow (r=a) vs Deep (r=a+z) Assume at z=637.12m, so r = 1.1a For shallow dynamic u = acosφ dλ For deep atmosphere u = r cosφ dλ For example, ϕ=45, u=400m/s, after one hour advection, the displacement has about 1 error in λ. 10 April 2014 PDEs 2014, H. Juang 6

7 Deep atmospheric equation in height spherical coordinates du uvtanφ + uw r r (2Ωsinφ)v + (2Ωcosφ)w + 1 p ρ r cosφ = F u Momentum where dv + u2 tanφ + vw r r + (2Ωsinφ)u + 1 p ρ r φ = F v dw u2 + v 2 r da = A t + u A r cosφ + v A r φ + w A r p = p n = ρ n R n T = ρ n n n ρ n R n ρ ρ t + ρu r cosφ + ρvcosφ r cosφ φ + ρr2 w r 2 r = F ρ T = ρ (2Ωcosφ)u + 1 p ρ r + g = F w n q n R n T = ρrt r = a + z u = r cosφ dλ v = r dφ Gas law Density 10 April 2014 PDEs 2014, H. Juang 7

8 From IEE ρe t + ρev + p V = ρq replace N e = q i e i = q i C Vi T = C V T i=1 N i=1 = ( C P R)T = h RT We have where N ρh t Combine IEE in h form with + ρhv dp = ρq C V = q i C Vi C P = q i C Pi R = q i R i i=1 N i=1 N i=1 ρ t + ρv = F ρ Specific value q i = ρ i ρ ρ = N n=1 ρ n We have dh κh p dp = Q h ρ F ρ we need dq i = F q i where h=cpt is an Enthalpy 10 April 2014 PDEs 2014, H. Juang 8

9 du uvtanφ + uw r r f sv + f c w + κh p dv dw + u2 tanφ r + vw r u2 + v 2 r + f s u + κh p f c u Deep Atmos Dyn in generalized coordinate Staniforth and Wood (2003) where d() = () t + λ () +φ () φ +ζ () 1 p r cosφ p r = F u r 1 r p φ p r = F v r φ + κh p f s = 2Ωsinφ ; f c = 2Ωcosφ ; g = g(r) ; p = ρκh p r + g = F w dh κh p dp = F h β t + β λ + β φ φ + βζ = F β ρ dq i = F q i ; β = ρr 2 cosφ r ; w = r t + λ r +φ r φ +ζ r 10 April 2014 PDEs 2014, H. Juang 9

10 da Angular momentum put = r cosφ du We have + ( u + 2Ωr cosφ) du = uvtanφ uw r r + f sv f c w κh p da = r cosφ ( * F u κh ) p A = r cosφ ( u + Ωr cosφ) dr cosφ = r cosφ du + ( u + 2Ωr cosφ) ( wcosφ vsinφ) 1 p r cosφ p r + F u r 1 p r cosφ p r + - r, in da = A t + u A r cosφ + v A A r φ +ζ β t + u r cosφ β v + r β β φ + = 0 βa t u r cosφ βa + v + r βa + βa φ = β da where β = ρr 2 cosφ r 10 April 2014 PDEs 2014, H. Juang 10

11 βa +! t ) = βr cosφ+ F u κh * p uβa +! r cosφ φ vβa r + βa 1! p r cosφ p r,. r - = r 2 cosφ r p + r2 cosφ p r + βr cosφf u pr 2 cosφ r = + p pr 2 cosφ r = + t r 2 cosφ r + pr 2 cosφ r p pr 2 cosφ r + βr cosφf u ρadv = ρr cosφf u dv r 2 cosφ r + βr cosφf u + p r ( ds p r ( T B ds Angular momentum conserved if top p=0 or r=constant thus the bottom is the only torque. 10 April 2014 PDEs 2014, H. Juang 11

12 Consider Kinetic energy from momentum equations u du u uvtanφ + u uw r r uf v + uf w + uκh s c p v dv w dw + v u2 tanφ r + v vw r w u2 + v 2 K t + λ K +φ K φ +ζ K = 1 ρ λ r + vf s u + v κh p wf c u Sum them together, we have βk t β t + β λ + β φ φ + βζ 1 p r cosφ p r = 0 r 1 r p φ p r = 0 r φ + w κh p Combine with mass and w equations as p r + wg = 0 p p r ( 1 p r ρ φ φ p r ( 1 r φ ρ w p r gw = 0 and w = r λ β K + p φ β K + p β K + p ρ ρ ρ φ β ρ λ β = p + ρ φ β φ + ρ ζ pr 2 cosφ r t β ρ λ = p r 2 cosφ r + p t t + r r r λ +φ φ +ζ β + p ρ φ φ β ρ βgw = p t ( ) K = 1 2 u2 + v 2 + w 2 β + p ρ ζ p β r ρ t r + p β ρ λ β + + ρ φ β φ + ρ ζ We have β r ρ t r βgw pr 2 cosφ r t βgw 10 April 2014 PDEs 2014, H. Juang 12

13 Consider geo-potential energy gdr = Φ or g = dφ gw = dφ dr dr = Φ t + Φ Φ Φ λ +φ φ +ζ Combine with mass equation We have βφ t + λ βφ Consider internal energy dc P T 1 dp ρ = dc VT + φ βφ + βφ φ + drt β t + β λ + β φ φ + βζ = 0 = βgw 1 dp ρ = dc VT d 1 ρ + p = 0 Apply mass equations in both total derivative terms, We have βc V T t + βc V T + φ βc V T φ + βc V T ) = p β (+ β t ρ λ (+ β ρ φ φ (+ β, + ρ ζ (. * ρ - 10 April 2014 PDEs 2014, H. Juang 13

14 Combine previous three red enclosed equations βk t βφ t βc V T t λ β K + p φ β K + p β K + p β β ρ ρ ρ ρ ρ λ β = p + φ r t + ρ φ β φ + ρ ζ + λ βφ + βc V T + φ βφ + βφ φ + φ βc V T φ βk + βc T V + βφ ( t t t = βgw + βc V T Integral globally with all BC, include dζdλdφ = - pr 2 cosφ r ) = p β (+ β t ρ λ (+ β ρ φ φ (+ β, + ρ ζ (. * ρ - r p ζt = 0 = 0 t ζ B +, ( t ζ T pr 2 cosφ r ( t pr 2 cosφ r t ζ B. 0dλdφ / βgw and BC give t ( ) β K + C V T + Φ dζdλdφ = t Total energy conserved ρ ( K + C V T + Φ) dv = 0 10 April 2014 PDEs 2014, H. Juang 14

15 Deep Atmos vs non- Hydro From Deep atmosphere, we require r changes with Tme, thus we need dw/ equaton And we need full curvature and Coriolis force terms to satsfy conservaton Thus, based on conservaton requirement, a deep atmospheric dynamic is a non- hydrosta)c dynamic. A non- hydrosta)c dynamics can be shallow or deep atmospheric dynamics. Both r and vertcal components of curvature and Coriolis force should be considered in deep atmosphere; and should not be considered in shallow atmosphere. 10 April 2014 PDEs 2014, H. Juang 15

16 du * + u* w r dv * + v* w r dw s * 2 m2 r f s v * + f c * w + κh p + f s u * + m 2 s * 2 Deep Atmos Dyn in spherical mapping generalized coordinates m 2 f c * u * r sinφ + κh p ρ * t + κh p 1 r 1 r * ρ * u + m 2 r p p r r p ϕ p r r ϕ p r * v * ρ + r m2 ϕ dh κh p = F u = F v +g = F w dp = F h + ρ* ζ = F * ρ Staniforth and Wood (2003) Juang (2014) NCEP Office Note dq i = F q i p = ρκh where d() = () t + λ () +ϕ () ϕ +ζ () = () t + m2 u * () r + m2 v * () r ϕ +ζ () f s = 2Ωsinφ ; f c * = 2Ωcos 2 φ ; g = g(r) ; κ = R C P ; γ = C P ; s * 2 C V ; ρ * = ρ r2 a 2 r = u *2 + v *2 10 April 2014 PDEs 2014, H. Juang 16

17 Start from continuity equation to have similar to shallow and hydrostatic system to mae as mass coordinates, Mass at give area can be obtained by integral vertical as Mass = φ 2 φ 1 λ 2 λ 1 ζ Top ζ sfc ρr 2 cosφ r dζ dλ dφ Then project this mass to earth surface pressure = Mass g Area = ρgr 2 cosφ r dζ dλ dφ a 2 cosφ dλ dφ We define it as coordinate pressure 10 April 2014 PDEs 2014, H. Juang 17

18 Previous integral can be deduced to be only in vertical as ζ TOP ρg r2 =!p ζsurface = ζ surface r a 2 dζ ζ TOP ζ surface!p dζ Thus, we have!p r2 r = ρg a 2 = ρ* g Put into continuity equation ρ * t ρ * u + m 2 r * * v * ρ + r m2 ϕ + ρ* ζ = 0 We have p t p u * + m 2 r p + ϕ v * r p ζ + = 0 10 April 2014 PDEs 2014, H. Juang 18

19 Since!p = ρ* g and ρ * > 0 Thus!p is monotone with vertical coordinate We can use it for coordinate definition and we can call it height weighted coordinate pressure or simply coordinate pressure So we can have!ˆp = Â + ˆB!p s + Ĉ h 1 + h h h 0 C pd /R d For opr compatibility, we use!ˆp = Â + ˆB!p s 10 April 2014 PDEs 2014, H. Juang 19

20 =K, last layer at the top ζˆ +1 ˆp +1 ˆr +1 u v w h p q ζˆ ˆp ˆr =1, the first layer next to ground For dynamics, all source terms are set to be zero! 10 April 2014 PDEs 2014, H. Juang 20

21 Vertical integral angular momentum principle, we have ζ T p κh p p p ζ r T (dζ = p r ζ S ζ S κh p p p κh p p r (dζ r ζ T = r gr 2 p a 2 + gr2 p r (dζ a 2 ζ S ζ T = g r3 p 3a 2 + p r 3 (dζ ζ S ζ T = g 3a 2 ζ S r3 p + r 3 p r3 = g r 3 p 3 p ζ T r ( 3a 2 ( dζ ζ S ( = ζ T g p 3a 2 r3 dζ gr T3 ζ S 3a 2 p T p ( ( dζ ( + gr 3 S p S 3a 2 10 April 2014 PDEs 2014, H. Juang 21

22 Discretize the fourth and last LHSs ζ T ζ S g 3a 2 r3 p + r 3 p r3 With P top zero, we have K =1 p dζ = ζ T g p 3a 2 r3 dζ gr T3 p T ζ S 3a 2 + gr 3 S p S 3a 2 ( ( ˆr 3 +1 ˆr 3 ) p + r 3 ˆp +1 ˆp + * - ), = ˆr S 3 p S let p = f ( ˆp +1, ˆp ) Then we have so p = f ˆp +1 ˆp +1 + f ˆp ˆp and f ˆp +1 + f ˆp =1 K =1 ( ˆr 3 +1 ˆr 3 ) f 3 ˆp + r +1 ˆp +1 + ( ˆr 3 +1 ˆr 3 ) f 3 ˆp r ˆp = ˆr S 3 p S 10 April 2014 PDEs 2014, H. Juang 22

23 For simplicity, we let f ˆp = f ˆp +1 = 1 2 so r 3 = 1 2 ˆr ˆr 3 since ˆr 3 = +1 ˆr 3 + 3! κh pg so r 3 = ˆr ! κh pg a 2 ( ˆp ˆp ) +1 a 2 (!ˆp!ˆp ) +1 remove layer value we have ˆr 3 +1 = ˆr a2 g κ h ˆ p ˆp +1 p so r 3 = ˆr a2 g 1 ˆp κ i h i ˆp i+1 i + 3 i=1 p i 2 a 2 g κ h ˆ p ˆp +1 p 10 April 2014 PDEs 2014, H. Juang 23

24 We have ˆr 3 +1 a 2 = ˆr 3 a + 3! κh 2 pg ( p!ˆ!ˆp ) +1 then transform to spectral for r to do derivative, then transform bac to grid space for nonlinear computing. du * dv * = u * w r = v * w r * + f s v * f s u f c * w 2 * m 2 s r κ h p sinφ κ h p a r p a g 6 a r p a ϕ g 6 ˆp ˆp +1 a p!ˆ!ˆp +1 r ˆp ˆp +1 a p!ˆ!ˆp +1 r ˆr 3 a a + ˆr a 2 ˆr 3 a ϕ a + ˆr a 2 10 April 2014 PDEs 2014, H. Juang 24

25 ! Select two forms of advection of PGF dp 1 p κh dp γ p = p H p κh p V H )+ p κh p ( p r w V H H r = p H p κh p V H ) p p κh ( p Continue discretization, we have ( ) r w V H H r ( ) dp = γ p 1!p κh H!p κh p V 1!p κh H ) γ p (!p κh p p p! = γ p H ( V H ) + 1 Δr V 3 H H Δr 3 1 Δr Δ ( V 3 H H r 3 ) + a2 Δ!w r 2 Δr!! = γ p + µ µ! + u *! v * - = γ p m 2 r + - r - ϕ -, λ 1 λ 2 3 ˆr + λ λ ˆr +1 + µ 1 µ 2 Δr 3 10 April 2014 PDEs 2014, H. Juang 25 ) ( p p κh p r w V H H r 3 ˆr µ + µ µ +1 2 ( ) 3 ˆr +1 µ r w V H H r + a2 ( ) ) (! Δ!w r 2 Δr! * u 1 u * ˆr 3 r 1 r +! u * u * +1 ˆr 3 +1 r r +1 +! v * 1 v * ˆr 3 r 1 r ϕ +! v * v * +1 r r +1 ˆr ϕ 0 2Δr / + a2 ) (! Δ!w r 2 Δr

26 So we use dp/ from momentum equation to thermodynamic equation, thus total energy will be conserved. dh so ) u * v * + = γ κ h m 2 r + ( * r ( + ϕ (, + m 2 * u 1 * r 1 u r dh κh p dp = 0 ( ˆr 3 + u * u * +1 ( ˆr 3 +1 r r +1 + v * 1 v * ( ˆr 3 r 1 r ϕ + v * v * +1 r r +1 ( ˆr 3 +1 ϕ 2Δr 3 Δ!w ( r 2 Δr + a2. + / where w = r2 a 2 w 10 April 2014 PDEs 2014, H. Juang 26

27 dw = s * 2 m2 r + m2 f u * κh c* p p r g = s * 2 m2 r + m2 f * c u * + g r2 Δp a 2 Δ p g dr 2 w and so ( ) 2 = 2rw 2 + r 2 dw = 2 r 2 w + m 2 rs * r 3 w = r2 a 2 w d w = w 2 2a2 r + r 2 3 m2 a 2 s* + m 2 r 2 a m 2 r 2 f * c u * + ga 2 r 4 Δp a 4 Δ p 1 ( f c * u * + g r 4 Δp a 4 Δ p 1 ( Since And BC r 2 w = 1 2 ˆr 2 +1ŵ ˆr 2 ŵ so w = 1 2 ŵ ŵ * * ŵ 1 = m 2 u 1 ˆr 1 ˆr 1 + v 1 ˆr 1 m2 ˆr 1 ϕ so ŵ1 = ˆr 1 m2 a u 1 * ˆr 1 a + v * ˆr 1 1 a ϕ Then we have w at all levels 10 April 2014 PDEs 2014, H. Juang 27

28 Linearize for SISL Collect discretized equations Linearization Matrixes for semi-implicit Add semi-lagrangian with semi-implicit 10 April 2014 PDEs 2014, H. Juang 28

29 To linearize, we define a base state start ˆp 01 = ; ˆr 01 = a = ; h o = C Pd T 0 ; T 0 = 300 ˆ p 0 = Â + ˆB ˆp 01 ; Δ p 0 = ˆp 0 ˆp 0+1 ; 0 = R d / C Pd Since p equals to p bar, so p r2 r = ρg a 2 = pg r κh So we can get all r at interface as ( ) a2 then use ˆp 0 ˆp 0+1 =!ˆp 0!ˆp 0+1 r 0 ˆr 0+1 = ˆr 0 + κ 0h 0 g ln ˆp 0 ˆ p 0+1 ( ) 2 ; p 0 = 1 2 ˆp 0 + ˆp 0+1 then and r 0 ( ) Δp 0 = ˆp 0 ˆp = 1 2 ˆr ˆr 0+1 ε 0 = r 0 a 10 April 2014 PDEs 2014, H. Juang 29

30 Summary of linear equations in matrix/vector short form!!!!! D p s Dt dp dh d w dd * L L L L = Π 1i D i * L = b 0 p s + Γ i p i = d 0 D * + Μ i w i = f 0 D * + Ζ i w i = n(n +1) a 2 ( Α i p i + Β i h i + e p s ) 10 April 2014 PDEs 2014, H. Juang 30

31 δd * n(n +1) αδt ( Α a 2 i δ p i + Β i δh i + e δ p ) s = S D* δ w +αδt ( b 0 δ p s Γ i δ p ) i = S w δh +αδt ( f 0 δd * Ζ i δ w ) i = S h δ p +αδt ( d 0 δd * Μ i δ w ) i = S p δ p s +αδtπ 1i δd i * = S p s To solve it, we put ps and p into w to get w as function of D Then put w as function of D into h and p, thus we get p, h, and ps are all function of D, then put them into D equation to solve D 10 April 2014 PDEs 2014, H. Juang 31

32 Hydrostatic IC Surface pressure Divergence/vorticity => U/V Use coordinate pressure to interpolation Use hydrostatic pressure as coordinate pressure and nonhydrostatic pressure Generate vertical velocity from w=dz/ equation 10 April 2014 PDEs 2014, H. Juang 32

33 From hydrostatic system, we have all p at interfaces as ˆp = Â + ˆB p s ( p!ˆ ) = (!ˆp ) +1 + a + ẑ +1 hydro hydro ( ) 3 ( a + ẑ ) 3 3a 2 pg κh The new nonhydrostatic coordinate pressure is From hydrostatic relation, we have ẑ +1 = ẑ + ( ˆp ˆp +1 ) κ h p g So coordinate pressure in hydrostatic system is!ˆ p = Â + ˆB!ˆp s p = 1 ( 2 ˆp + ˆp +1 ) So interpolate u, v, h, and q from hydro to new coordinates 10 April 2014 PDEs 2014, H. Juang 33

34 Remaining p and w at deep atmospheric system From coordinate pressure definition ˆr 3 = +1 ˆr 3 + 3! κh a 2 (!ˆp!ˆp ) +1 pg and hydrostatc relaton ˆr +1 = ˆr + ( ˆp ˆp +1 ) κ h p g combine both by eliminating h/(pg), we have! ˆp ˆp +1 = 3 ˆr +1 ˆr a 2 (!ˆp ˆr 3 +1 ˆr 3!ˆp ) +1 can obtain p from top to surface p = 1 ( 2 ˆp + ˆp +1 ) 10 April 2014 PDEs 2014, H. Juang 34

35 Get vertical velocity w in deep atmospheric system ˆr 3 = +1 ˆr 3 + 3! κh a 2 ( ˆp ˆp ) from +1 Do d/ pg We have where! dp! = γ p H V H ŵ+1 = ŵ + κ h p g ΔB d ˆp s K = m 2 ˆp ˆp +1 =1 ( ) + 1 d ˆp s ( ) u * v * r + ( r ( ϕ ( Δ p κ h γ p 2 g Δr V 3 H H Δr3 1 Δr Δ V 3 H H r3 dp ( ) + a2 Δ w r 2 Δr and ŵ1 = m2 ˆr 1 a u 1 * ˆr 1 a + v * ˆr 1 1 a ϕ w = 1 2 ŵ ŵ 4/10/14 Henry Juang 35

36 Summary Deep atmospheric dynamics is discretized in generalized hybrid coordinates Semi-implicit semi-lagrangian is used Initial condition with hydrostatic system is used It is suitable for very high resolution and coupling with space model Details can be found in Juang 2014, NCEP office note 477 Coded and still under testing. 10 April 2014 PDEs 2014, H. Juang 36