Assignments 3 Question#1:Hadley cell under different forcing 1.15 日E(φ，) =1- 1.1 o H湘d-H8u(1282}讨论了当外1汗e的经向分布子二次1计箭多项式，即 1.05 日E(4, 2-1-子uA6m1+A(后-方微hx 路德环流内的 ⊙ 1 风场、裂辰场、环变的卫句充出等将在样狮度和许力当追的凿度而变化 灯集疼力强这的空间公布减为 9(4,) 0.95 0. =1-Au(sin'中-2+4（元-2 0.9 1.请推导出合德乘环源内的高空风坯和塞直T均位场 婿灯何研结厨分布： 日 0.85 日E(中，) Θ。 1-△a(in3o-+A(-》 2,同样利用小年要翻设，清港异出环讯的空间范正中用的表这式。如黑设 rgH 0.8 2清分员面出当△H=3和△H=16时.与He-ou的 ---Held 况相比，中H怎样联「而变作， -New 3.逆做日：在出情况下，近地F风场的分布有怎华查化。 0.75 0.2 0.4 0.6 0.8 sinφ 授课教师：张洋 1
授课教师：张洋 1 Assignments 3 Question#1: Hadley cell under different forcing 0 0.2 0.4 0.6 0.8 1 0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 sin E/o Held New ⇥E(, z) ⇥o = 1 2 3 HP2(sin ) + v( z H 1 2 ) ⇥E(, z) ⇥o = 1 H(sin3 1 4 ) + v( z H 1 2 )
Assignments 3 Question#1:Hadley cell under different forcing Angular momentum: 1.15 日E(φ，) =1- 2 [u=2a in2Φ 1.1 三UM CoSΦ 1.05 Thermal wind relation: 1 fu(0）-wol+tanu2()-2o1= gH∂© u0.95 a日。0b 0.9 0.85 日E(中，) (0)-(φ) D2a2 sin Θ。 1-△a63-+A(疗-》 日o 2gH cos2 0.8 ---Held -New 0.75 Need to know 白(0) 0.2 0.4 0.6 0.8 sinφ 日. 授课教师：张洋 2
⇥˜ (0) ⇥o 授课教师：张洋 2 Assignments 3 Question#1: Hadley cell under different forcing [u] = a sin2 cos ⌘ UM ˜ - vertically averaged potential temperature o - reference potential temperature f[u(H) u(0)] + tan a [u2(H) u2(0)] = gH ao ⇥˜ ⇥ 授教：洋 19 Held-Hou model -Thermal wind relation ! Thermal wind relation: = p fu + u2 tan a = 1 a ⇥ ⇥ From steady state momentum equation f[u(H) u(0)] + tan a [u2(H) u2(0)] = 1 a ⇥ ⇥ At z=H and Z=0 [(H) (0)] ⇥ z = g o Hydrostatic balance: ⇥(H) ⇥(0) H = g ˜ o Vertical integral from 0 to H ! Angular momentum: Thursday, September 30, 2010 [u] = a sin2 cos ⌘ UM ˜ - vertically averaged potential temperature o - reference potential temperature f[u(H) u(0)] + tan a [u2(H) u2(0)] = gH ao ⇥˜ ⇥ 授教：洋 19 Held-Hou model -Thermal wind relation ! Thermal wind relation: = p fu + u2 tan a = 1 a ⇥ ⇥ From steady state momentum equation f[u(H) u(0)] + tan a [u2(H) u2(0)] = 1 a ⇥ ⇥ At z=H and Z=0 [(H) (0)] ⇥ z = g o Hydrostatic balance: ⇥(H) ⇥(0) H = g ˜ o Vertical integral from 0 to H ! Angular momentum: Thursday, September 30, 2010 [u] = a sin2 cos ⌘ UM ˜ - vertically averaged potential temperature o - reference potential temperature f[u(H) u(0)] + tan a [u2(H) u2(0)] = gH ao ⇥˜ ⇥ 授教：洋 19 Held-Hou model -Thermal wind relation ! Thermal wind relation: = p fu + u2 tan a = 1 a ⇥ ⇥ From steady state momentum equation f[u(H) u(0)] + tan a [u2(H) u2(0)] = 1 a ⇥ ⇥ At z=H and Z=0 [(H) (0)] ⇥ z = g o Hydrostatic balance: ⇥(H) ⇥(0) H = g ˜ o Vertical integral from 0 to H ! Angular momentum: Thursday, September 30, 2010 [u] = a sin2 cos ⌘ UM f[u(H) u(0)] + tan a [u2(H) u2(0)] = gH ao ⇥˜ ⇥ Set u(0) = 0 2⇥ sin ⇥a sin2 cos + tan a ⇥2a2 sin4 cos2 = gH ao ⇥˜ ⇥ Integrate with respect to ˜ (0) ˜ () o = ⇥2a2 2gH sin4 cos2 授教：洋 20 Held-Hou model -Thermal wind relation ! Thermal wind relation: ! Angular momentum: Thursday, September 30, 2010 [u] = a sin2 cos ⌘ UM f[u(H) u(0)] + tan a [u2(H) u2(0)] = gH ao ⇥˜ ⇥ Set u(0) = 0 2⇥ sin ⇥a sin2 cos + tan a ⇥2a2 sin4 cos2 = gH ao ⇥˜ ⇥ Integrate with respect to ˜ (0) ˜ () o = ⇥2a2 2gH sin4 cos2 授教：洋 20 Held-Hou model -Thermal wind relation ! Thermal wind relation: ! Angular momentum: Thursday, September 30, 2010 Need to know 0 0.2 0.4 0.6 0.8 1 0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 sin E/o Held New ⇥E(, z) ⇥o = 1 2 3 HP2(sin ) + v( z H 1 2 ) ⇥E(, z) ⇥o = 1 H(sin3 1 4 ) + v( z H 1 2 )
Assignments 3 Question#1:Hadley cell under different forcing 1.15 Temperature should be continuous at the edge: 1.1 e-1-△B(ino)+A(7- 白（中H)=日E(pH） 1.05 Hadley cell does not produce net heating but just carry heat poleward over the rφH 白cos od 0.9 0.85 ⊙E,=1-△(sim3b-+△，（月 日。 Assume small o,sinΦ~Φ 0.8 ---Held -New 15gH△H 0150 0.2 0.4 0.6 0.8 OH= sin中 822a2 6(0)6E(0) 22a2 6E(0)1 日o 日o Θ。16 △H 授课教师：张洋 3
H = 15 8 gHH ⌦2a2 ⇥˜ (0) ⇥o ⇡ ⇥˜ E(0) ⇥o H3 H + ⌦2a2 2gH 4 H = ⇥˜ E(0) ⇥o 1 16 H3 H 授课教师：张洋 3 Assignments 3 Question#1: Hadley cell under different forcing ˜ (H) = ˜ E(H) Z H 0 ˜ cos d = Z H 0 ˜ E cos d 授教：洋 23 Held-Hou model -Extent of Hadley Cell ! Temperature should be continuous at the edge: ! Extent of Hadley Cell , with following considerations: H ! Hadley cell does not produce net heating but just carry heat poleward over the extent of Hadley: Cell Thursday, September 30, 2010 ˜ (H) = ˜ E(H) Z H 0 ˜ cos d = Z H 0 ˜ E cos d 授教：洋 23 Held-Hou model -Extent of Hadley Cell ! Temperature should be continuous at the edge: ! Extent of Hadley Cell , with following considerations: H ! Hadley cell does not produce net heating but just carry heat poleward over the extent of Hadley: Cell Thursday, September 30, 2010 n Hadley cell does not produce net heating but just carry heat poleward over the Assume small , sin ⇠ 0 0.2 0.4 0.6 0.8 1 0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 sin E/o Held New ⇥E(, z) ⇥o = 1 2 3 HP2(sin ) + v( z H 1 2 ) ⇥E(, z) ⇥o = 1 H(sin3 1 4 ) + v( z H 1 2 )
Assignments 3 根据小角度假设(sinb~中，cosp~1) 垂直平均之后的外力强迫为： 6@=1-daim3o-} 8, e。 e. 2-△u 垂直平均位温场随纬度分布为： 6(0)-6(o)22a2sin4p 60) 22a2 日a 2gH cos2 a。= g日0 O(中)=OE(DH)连续条件： 60y_02u2 2gH够s 6s0) △u经 60 6r(0)_ 22a2 9。 e. 9。 2g日i-4p% 0 scos地 守恒条件： 22a2 6。- 2-△rg)o 60)_ 6s(0)n2a2 e。 0. 授课教师：张洋 4
授课教师：张洋 4 Assignments 3 [u] = a sin2 cos ⌘ UM f[u(H) u(0)] + tan a [u2(H) u2(0)] = gH ao ⇥˜ ⇥ Set u(0) = 0 2⇥ sin ⇥a sin2 cos + tan a ⇥2a2 sin4 cos2 = gH ao ⇥˜ ⇥ Integrate with respect to ˜ (0) ˜ () o = ⇥2a2 2gH sin4 cos2 授教：洋 20 Held-Hou model -Thermal wind relation ! Thermal wind relation: ! Angular momentum: Thursday, September 30, 2010 连续条件： 守恒条件：
Assignments 3 Question#1:Hadley cell under different forcing 1.15 1.1 9-1-B+A(后- 80 1.05 --AH=1/3,Held-Hou 70 -AH=1/3,New ---AH=1/6,Held-Hou 10.95 60 -AH=1/6,New 0.9 50 0.85 ⊙E,=1-△(sin3b-)+△，（月 (6ap) 5gH△H 日o 40 中H.= 3 22a2 0.8 ---Held -New 30 0150 0.2 0.4 0.6 0.8 sin 20 10 15gH△H H台 822a2 0.5 1 1.5 2 r 授课教师：张洋 5
0 0.5 1 1.5 2 0 10 20 30 40 50 60 70 80 r H (deg) H=1/3,Held−Hou H=1/3,New H=1/6,Held−Hou H=1/6,New 授课教师：张洋 5 Assignments 3 Question#1: Hadley cell under different forcing H = ✓5 3 gHH ⌦2a2 ◆1/2 0 0.2 0.4 0.6 0.8 1 0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 sin E/o Held New ⇥E(, z) ⇥o = 1 2 3 HP2(sin ) + v( z H 1 2 ) ⇥E(, z) ⇥o = 1 H(sin3 1 4 ) + v( z H 1 2 ) H = 15 8 gHH ⌦2a2
Held-Hou model -Surface winds From the zonal momentum equation MOSPHERIC SCIENCES VOLUME 37 0=-.(vw)+f元+4wtan 0 Bu 0 u2 tang 1Φ 0=-7.(vv)-fu- a 08 dvv 8/8v × (1) 60 0=-V·(v8)-(⊙-⊙ε)r1+ 1 (u日cosp)dz= 0E-⊙ 0=-7v Jo a cos o 0o 入 8Φ =g0/Θ0 02 Then,mass flux V can be solved Similarly,we have with boundary conditions the momentum flux, du ov 80 at z=H:w=0; =0 H bo uudz VUm at z=0: w=0; =0； .(1a) 0z =Cu; =C 0z 又木子X甲·瓜干
1 a cos2 @ @ cos2 Z H 0 uvdz! = Cu(0) 1 ⇥o Z H 0 v⇥dz ⇡ V V 1 H Z H 0 1 a cos @ @(v⇥ cos )dz = ⇥˜ E ⇥˜ ⌧ 授课教师：张洋 6 Held-Hou model -Surface winds n From the zonal momentum equation Z H 0 uvdz ⇡ V Um Then, mass flux V can be solved. Similarly, we have the momentum flux
Held-Hou model -Surface Winds From thermodynamic equation: 1 H日oJ oaw-863-品5ag-aeg密 ac0sp∂0 Γ2gH0) eo=2学rs-+后的d H Θ0J0 T ΦH Then,mass flux V can be solved. H uudz VUm 0 Cu0)=- △v T ΦH 授课教师：张洋 7
授课教师：张洋 7 Held-Hou model -Surface Winds n From thermodynamic equation: Hadley ω ∂Θ ∂z ≈ ΘE − Θ τ (16) 1 Θ0 ∂Θ ∂z ≈ ∆v H (17) ω ≈ ΘE − Θ Θ0 H τ∆v (18) φ = 0 ω ≈ H ∆vτ (∆Hφ3 H + 1 2 Ω2a2 gH φ4 H) ≈ 1 16( 15 8 ) 3 ∆4 Hr3 (19) 1 HΘ0 ! H 0 1 acosφ ∂ ∂φ(vΘcosφ)dz = Θ¯ e − Θ¯ τΘ0 = 1 τ ( 1 16( 15 8 ) 3 ∆4 Hr3 − ∆Hφ3 + 1 2 Ω2a2 gH φ4 ) (20) φ 1 Θ0 ! H 0 (vΘ)dz = ! φ 0 Ha τ ( 1 16( 15 8 ) 3 ∆4 Hr3 − ∆Hφ3 + 1 2 Ω2a2 gH φ4 ) dφ (21) 1 Θ0 ! H 0 (vΘ)dz = Har4∆5 H τ 1 16( 15 8 ) 4 [ φ φH − 4( φ φH ) 4 + 3( φ φH ) 5 ] (22) 1 Θ0 ! H 0 vΘdz ≈ V ∆v (23) ! H 0 vudz ≈ V uM (24) 1 acos ∗ 2φ ∂ ∂φ(cos2 φ ! H 0 vudz) = −Cu(0) (25) Cu(0) = 1 a ∂ ∂φ(V uM) = 1 a ∂ ∂φ( V ∆v ∆v uM) (26) uM ∆v = Ωaφ2 ∆v (27) 6 Hadley ω ∂Θ ∂z ≈ ΘE − Θ τ (16) 1 Θ0 ∂Θ ∂z ≈ ∆v H (17) ω ≈ ΘE − Θ Θ0 H τ∆v (18) φ = 0 ω ≈ H ∆vτ (∆Hφ3 H + 1 2 Ω2a2 gH φ4 H) ≈ 1 16( 15 8 ) 3 ∆4 Hr3 (19) 1 HΘ0 ! H 0 1 acosφ ∂ ∂φ(vΘcosφ)dz = Θ¯ e − Θ¯ τΘ0 = 1 τ ( 1 16( 15 8 ) 3 ∆4 Hr3 − ∆Hφ3 + 1 2 Ω2a2 gH φ4 ) (20) φ 1 Θ0 ! H 0 (vΘ)dz = ! φ 0 Ha τ ( 1 16( 15 8 ) 3 ∆4 Hr3 − ∆Hφ3 + 1 2 Ω2a2 gH φ4 ) dφ (21) 1 Θ0 ! H 0 (vΘ)dz = Har4∆5 H τ 1 16( 15 8 ) 4 [ φ φH − 4( φ φH ) 4 + 3( φ φH ) 5 ] (22) 1 Θ0 ! H 0 vΘdz ≈ V ∆v (23) ! H 0 vudz ≈ V uM (24) 1 acos ∗ 2φ ∂ ∂φ(cos2 φ ! H 0 vudz) = −Cu(0) (25) Cu(0) = 1 a ∂ ∂φ(V uM) = 1 a ∂ ∂φ( V ∆v ∆v uM) (26) uM ∆v = Ωaφ2 ∆v (27) 6 Z H 0 uvdz ⇡ V Um Then, mass flux V can be solved. Cu(0) = − Ω ∆v Har5∆6 H τ 1 16( 15 8 ) 5 [3( φ φH ) 2 − 24( φ φH ) 5 + 21( φ φH ) 6 ] (28) 7
Held-Hou model -Surface Winds Cu0=-是Haa9PP-24品r+21号月 △v 168 D 地面东西风分布（西风为正） 0.6 0.5 Held-Hou revision 0.4 0.3 0.2 0.1 -0.1 -0.2 -0.3 -0.4 0.10.20.3 0.4 0.5 0.6 0.7 0.8 0.9 phi/phiH 授课教师：张洋 8
授课教师：张洋 8 Held-Hou model -Surface Winds Cu(0) = − Ω ∆v Har5∆6 H τ 1 16( 15 8 ) 5 [3( φ φH ) 2 − 24( φ φH ) 5 + 21( φ φH ) 6 ] (28) 7
