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