Thermocline problem
313
0.98
1
1.02
1.04
1.06
1.08
1.1
1.12
1.14
288
296
304
312
320
328
336
344
60N
50N
40N
h
2
H
2
___
φ φ
φ
φ
Figure 13.4. Plot of the solution h2(φ, θ) as in (13.62) versus φ for different values of θ.T h e
value of α/(γH
2
2 )=0.5 and the domain is [φW ,φE] = [286, 350] and [θ1,θ0]=[40, 70].
Now consider the changes in the potential vorticity of layer 2, i.e.,
q 2 =
sin θ
h 2
,
(13.66)
along curves of constant pressure p 2 . These curves coincide with streamlines
because (with (13.52a-b)),
u 2 ·∇p 2 =
u 2
v 2
.
1
cos θ
∂p2
∂φ
∂p2
∂θ
=0,
(13.67)
and hence the velocity vector is tangent to curves of constant pressure. Elimination of the pressure in (13.52a-b) gives the vorticity equation
v 2 cos θ =sinθ
∂w 2
∂z
.
(13.68)
Integration over the layer (with thickness h 2 ) gives, with (13.64c),
z3
z2
v 2 cos θdz = h 2 v 2 cos θ =sinθ(
Dz 2
dt
−
Dz 3
dt
)=sinθ
Dh 2
dt
,
(13.69)
and hence with D(sin θ)/dt = v 2 cos θ, it is found that
Dq 2
dt
=
1
h 2
D(sin θ)
dt
−
sin θ
h 2
2
Dh 2
dt
=0.
(13.70)