Antarctic Circumpolar Current
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14.2.1. Flat bottom case
In the flat bottom case, we find from (14.4a) that the vertical velocity w =0at
z = −D. Integration of the equations (14.3) over depth (from −D to 0)gives
−2Ω¯ v sin θ = −
1
r 0 ρ 0 cos θ
∂ ¯
p
∂φ
+
τ φ
ρ 0
−
τ
φ
b
ρ 0
+ ¯
F
φ ,
(14.5a)
2Ω¯ u sin θ = −
1
r 0 ρ 0
∂ ¯
p
∂θ
+
τ θ
ρ 0
−
τ θ
b
ρ 0
+ ¯
F
θ ,
(14.5b)
0=
∂ ¯
u
∂φ
+
∂(¯ v cos θ)
∂θ
,
(14.5c)
where the barred quantities indicate vertically integrated quantities, i.e.,
¯
u =
0
−D
udz
and τ b is the bottom shear stress.
In a zonally periodic channel with τ φ = τ φ (θ) and τ θ =0 , we have seen in
Example 13.3 that there exists solutions which are independent of φ with ¯
v =0 .
For these solutions, the zonal momentum balance (14.5a), provides two possibilities: (i) the wind-stress forcing is compensated for by bottom friction or (ii) it is
compensated by lateral friction. In case (ii) (case (i) will be subject exercise 14.1)
assume that the lateral friction is proportional to the lateral viscosity A H and is
of the form A H ∇ 2 u.I fU indicates a characteristic zonal velocity, then it follows
from a balance between the wind-stress term (τ φ /ρ 0 ) and the lateral friction term
( ¯
F φ )that
Ex. 14.1
τ 0
ρ 0
≈
A H UD
r 2
0
→ U =
τ 0 r 2
0
ρ 0 DA H
,
(14.6)
and hence the transport Φ scales as
Φ=
θ2
θ1
r 0 ¯
udθ → r 0 DU =
τ 0 r 3
0
ρ 0 A H
.
(14.7)
◮
Example 14.1: Channel transport: flat bottom
Consider a channel with θ 1 =65 ◦ Sandθ 2 =55 ◦ S, for which the flow is forced
by the wind stress
τ
φ (θ)=τ 0 sin
(θ − θ 1 )
(θ − θ 1 )
; τ
θ =0.