Space Science: Atmospheres
Part- 6b
Viscosity + Eddy Mixing
Planetary Boundary Layer
Ekman Spiral
Spin Down Time
Reminder
k' x (-"pr )
r
= - f# v
Steady State
Geostrophic $ large scale motion
Near surface: we need to add in
drag forces
Result:
Winds near surface are often to
the left of the winds aloft in
northern henispshere
This defines the ‘boundary layer’
Planetary Boundary Layer
Surface is not smooth on
some scale
Scale of ‘roughness’ differs
over mountains, plains, cities
Typical average ~1km (a fraction of H)
treat turbulence induced as a viscosity
Ekman number not negligible
E-1 = Re = L V / Κ
; Κ = eddy viscosity
Viscosity
transport of turbulence (momentum)
between adjacent layers of fluid
Molecular--primarily above homopause
Eddy --below
No Coriolis Force (Equatorial)
u
Very Near Surface
z
Stress ρK[∂ u/∂ z]
~ const
Momentum Flux due to turbulence/ mixing
" = #K(du/dz)
" is a stress (units of pressure)
Reach a steady state with
Constant Stress between layers
but length/ mixing scale changes near the surface
K~Lv
L $ % z near surface; % von Karman const.
Ekman Spiral
Include Coriolis (assume curvature small)
viscosity and pressure gradients
"u
1 "p
"2
= 0 = fv #
+K 2u
"t
$ "x
"z
"v
1 "p
"2
= 0 = # fu #
+K 2 v
"t
$ "y
"z
2 – D Flow, but coupled vertically!
!
Boundary conditions
u(0) = 0 = v(0)
u (z $ # ) = u g , v(z $ # ) = v g
Geostrophic values from
1 !p
f vg =
% !x
1 !p
f ug = "
% !y
Use geostrophic solutions
"2
K 2 u = #f(v # v g )
"z
"2
K 2 v = f(u # u g )
"z
Trick : Define
V $ u + iv
Multiply v equation by i and
add the two equations
"2
K 2 (u + i v) = i f [ (u - u g ) + i(v # v g )]
"z
"2
K 2 (u + i v) = i f [ u + iv - (u g + iv g )]
"z
OR
"2
V = % (V # V g )
2
"z
% $ if/K
Solution (cont)
V "V g = a e
1)
2)
#1/2z
+ be
"# 1 / 2 z
V(0) = 0
V ($) = V g
# 1/2
3)
% i f (1/ 2
1+ i
1/ 2
= ' * = + (1+ i) : [i =
]
&K)
2
% f (1/ 2
+ = ' *
& 2K )
V = V g + a e+(1+ i)z + b e+(1+i)z
,
Use
(2)
V = V g + b e+(1+ i)z
Use
(1)
0 = Vg - b
b = Vg
,
V = V g [ 1 - e -+z e i+z ]
Analytic Solution!
decay and periodic terms
Solution (cont)
Need to go back to u, v
For simplcity : u g = u g
(V = u + iv)
vg = 0
Solutions!
u = ug [ 1 - e -"z cos "z]
v = ug e -"z sin "z
Ekman was actually an oceanographer
Example " = [ f /2K]1/ 2
# = 30 o Latitude
then
f = 7 $ 10 -5 /s
If
K = 100 m2 /s
" -1 = 1.6 km
" -1 = Thickness of Planetary Boundary Layer
Ekman Spiral
Solutions
at z >> γ-1
u=ug, v=0
u/ug
v/vg
Ekman Spiral
Wind speed and direction
vs. altitude
v (m/s)
z in meters
u (m/s)
Dashed: Measurements
Solid: Analyitc solution
mid latitude
Force Balance
If at 10 km, Northern hemisphere
1000 mb
1005 mb
p
ug
Cor.
1010 mb
Below ~ 1 km
1000 mb
v
v
p
1005 mb
1010 mb
ug
fric
cor
!
Surface winds in the Northern Hemisphere
tend to blow to the left of winds aloft.
Ocean Layer Driven by
Wind Shear (Ekman)
1. Wind direction; 2. Forcing
3. Actual Flow;
4. Geostrophic flow
In ocean waves and Turbulence disrupt
Ekman ‘Pumping’
The coupling of the geostrophic
layer (often circular motion) to
the surface through the planetary
boundary layer gives a spin down
time for the circular motion in the
troposphere
Not too hard to show
note: boundary layer size ~ γ-1
Spin down time( interaction with
the surface
τs ~ 2(Hγ) f-1
Mid latitudes ~ days
Summary
•
•
•
•
Eddy Viscosity
Planetary Boundary Layer
Ekman Spiral
Spin Down Time