&会品 Held-Hou model -Angular momentum Review [M=(acosφ+[u)acosφ At the equator,as the parcels rise from the surface,where the flow is weak,we assume that the zonal flow is zero there. 2 in-o [u=2a 三UM cos o acoSΦ Then,what is the UM at 10,20,30 degree? Answers:14,57,134 m/s,respectively Combined with the weak surface flow,this indicates strong vertical shear of the zonal wind. 授课教师：张洋7
M = (⌦a cos + u)a cos [u] = ⌦a sin2 cos ⌘ UM 授课教师：张洋 7 Held-Hou model -Angular momentum n At the equator, as the parcels rise from the surface, where the flow is weak, we assume that the zonal flow is zero there. ⌦ a cos a [ ] [ ] Then, what is the UM at 10, 20, 30 degree? Answers: 14, 57, 134 m/s, respectively Combined with the weak surface flow, this indicates strong vertical shear of the zonal wind. Review
&会易 Held-Hou model -Temperature distribution Review Angular momentum: sin2o 4=acos中 UM Thermal wind relation: fu(D)-uo1+tan°2(H)-2o1= gH∂g a日。ab ©(0)-Θ（Φ） 22a2sin40 o 2gH cos2 o Conservation of angular momentum and the maintenance of thermal wind completely determine the variation of temperature within the Hadley Cell 授课教师：张洋8
[u] = ⌦a sin2 cos ⌘ UM f[u(H) u(0)] + tan a [u2(H) u2(0)] = gH a⇥o @⇥˜ @ ⇥˜ (0) ⇥˜ () ⇥o = ⌦2a2 2gH sin4 cos2 授课教师：张洋 8 Held-Hou model -Temperature distribution n Thermal wind relation: n Angular momentum: Conservation of angular momentum and the maintenance of thermal wind completely determine the variation of temperature within the Hadley Cell ! Review
&会 Held-Hou model -Extent of Hadley Cell Review ΦH ΦH 白cosφd0= 白cos od0 0 Radiative equilibrium temperature 百FROM EQ.l2) 日(，=1- anA(mo+a(后-司 △H fractional temperature difference between equator and pole 百（中H)=日EH) △u fractional temperature difference between ground and top Psecond Legendre polynomial,()1) Vertical average: ⊙eo,习=1-号△aPB(sin0) 2 EQUATOR POLE LATITUDE 授课教师：张洋 9
授课教师：张洋 9 Held-Hou model -Extent of Hadley Cell ⇥˜ (H) = ⇥˜ E(H) Z H 0 ⇥˜ cos d = Z H 0 ⇥˜ E cos d H - fractional temperature di↵erence between equator and pole ⇥E(, z) ⇥o = 1 2 3 HP2(sin ) + v( z H 1 2 ) P2 - second Legendre polynomial, P2(x) = 1 2 (3x2 1) ⇥˜ E(, z) ⇥o = 1 2 3 HP2(sin ) n Radiative equilibrium temperature Vertical average: v - fractional temperature di↵erence between ground and top Review
&会 Held-Hou model -Extent of Hadley Cell Review Assume small,sinp~Φ ΦH ΦH 白cos odΦ= 白cos od0 0 (0)E(0) 5gH△ 1822a2 百[FROM EQ.l2) 1/2 d(pH)=ΘH) 2e /2 set R then,φH= R EQUATOR POLE LATITUDE 授课教师：张洋 10
Assume small , sin ⇠ 授课教师：张洋 10 Held-Hou model -Extent of Hadley Cell ⇥˜ (H) = ⇥˜ E(H) Z H 0 ⇥˜ cos d = Z H 0 ⇥˜ E cos d H = ✓5 3 gHH ⌦2a2 ◆1/2 set R = gHH ⌦2a2 , then, H = ✓5 3 R ◆1/2 ⇥˜ (0) ⇥o ⇡ ⇥˜ E(0) ⇥o 5 18 gH2 H ⌦2a2 Review
&会 Held-Hou model -Strength of Hadley Cell Thermodynamic equation at equator and steady state: DΘ ΘE-Θ ∂0、ΘE-Θ Dt T W02 T If the static stability is mostly determined by the forcing instead of meridional circulation: 1∂日 △v Θ。∂z H H ΘE-Θ 5g△2HH2 ≈ Θo△yT 18a2T22△V Using mass continuity: gH032△2 H Characteristic overturning time a23T△V Td w scale can be estimated. 授课教师：张洋11
授课教师：张洋 11 Held-Hou model -Strength of Hadley Cell 1 ⇥o @⇥ @z ⇡ V H n If the static stability is mostly determined by the forcing instead of meridional circulation: n Thermodynamic equation at equator and steady state: w ⇡ H ⇥oV ⇥E ⇥ ⌧ = 5g2 HH2 18a2⌧⌦2V v ⇠ (gH) 3/2 5/2 H a2⌦3⌧V ⌧d = H w Characteristic overturning time scale can be estimated. w @⇥ @z ⇡ ⇥E ⇥ ⌧ D⇥ Dt = ⇥E ⇥ ⌧ n Using mass continuity: