Presentation Slides for Chapter 4 of Fundamentals of Atmospheric Modeling 2 nd Edition
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Transcript of Presentation Slides for Chapter 4 of Fundamentals of Atmospheric Modeling 2 nd Edition
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Presentation Slides for
Chapter 4of
Fundamentals of Atmospheric Modeling 2nd Edition
Mark Z. JacobsonDepartment of Civil & Environmental Engineering
Stanford UniversityStanford, CA [email protected]
March 10, 2005
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Spherical Horizontal Coordinates
iλkrRecosϕ
λeReReϕ
jϕ
Fig. 4.1
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West-east and south-north increments (4.1)
Spherical Coordinate Conversions
dx = Recosϕ( )dλe dy=Redϕ
Example 4.1 dλe = 5o = 5o x / 180o=0.0873 radd = 5o =0.0873 rad = 30 oN
--> dx = (6371 km)(0.866)(0.0873 rad) = 482 km--> dy = (6371 km)(0.0873 rad) = 556 km
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Spherical coord. total and horizontal velocity vectors (4.2)
Spherical Coordinate Conversions
Scalar velocities (4.3)
v=iλu+jϕv+krw vh =iλu+jϕv
u =dxdt
=Recosϕdλedt
v=dydt
=Redϕdt
w =dzdt
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Gradient operator in spherical coordinates (4.4)
Spherical Coordinate Conversions
Dot product of gradient operator with velocity vector (4.5)
∇ =iλ1
Recosϕ∂
∂λe+jϕ
1Re
∂∂ϕ
+kr∂∂z
∇ •v= iλ1
Recosϕ∂
∂λe+jϕ
1Re
∂∂ϕ
+kr∂∂z
⎛
⎝ ⎜
⎞
⎠ ⎟ • iλu+jϕv+krw( )
=1
Recosϕ∂u∂λe
+iλu1
Recosϕ∂iλ∂λe
+iλv1
Recosϕ
∂jϕ∂λe
+iλw1
Recosϕ∂kr∂λe
⎛
⎝ ⎜
⎞
⎠ ⎟
+1Re
∂v∂ϕ
+jϕu1Re
∂iλ∂ϕ
+jϕv1Re
∂jϕ∂ϕ
+jϕw1Re
∂kr∂ϕ
⎛
⎝ ⎜
⎞
⎠ ⎟
+∂w∂z
+kru∂iλ∂z
+krv∂jϕ∂z
+krw∂kr∂z
⎛
⎝ ⎜
⎞
⎠ ⎟
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Spherical Coordinate Conversions
Fig. 4.2a,b
ΔλeRe cosϕRe cosϕ W
E
Δλei λ
i λi λ+Δi λΔi λ
Δx
Top view
ϕReRe
N
S -kr Δi λ cosϕΔi λjϕ Δi λ sinϕ ϕ
Side view
Δiλ =jϕ Δiλ sinϕ −kr Δiλ cosϕ
From Fig. 4.2a (4.6)Δiλ = iλ Δλe =Δλe
From Fig. 4.2b (4.7)
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Spherical Coordinate Conversions
Dot product of gradient operator and velocity vector (4.5)
∂iλ∂λe
≈jϕΔλesinϕ−krΔλecosϕ
Δλe≈jϕ sinϕ−kr cosϕ
Substitute (4.6) into (4.7), divide by Δλe (4.8)
∇ •v=1
Recosϕ∂u∂λe
+1
Recosϕ∂∂ϕ
vcosϕ( )+1
Re2
∂∂z
wRe2
( )
Δiλ =jϕ Δiλ sinϕ −kr Δiλ cosϕ
From Fig. 4.2a (4.6)Δiλ = iλ Δλe =Δλe
From Fig. 4.2b (4.7)
∇ •v=1
Recosϕ∂u∂λe
+iλu1
Recosϕ∂iλ∂λe
...⎛
⎝ ⎜
⎞
⎠ ⎟
Substitute (4.8) and other terms into (4.5) (4.10)
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Assume Re constant (4.11)
Spherical Coordinate Conversions
∇ •v=1
Recosϕ∂u∂λe
+1
Recosϕ∂∂ϕ
vcosϕ( )+∂w∂z
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Inertial Reference FrameInertial reference frame
Reference frame at rest or at constant velocity, such as one fixed in space
Noninertial reference frameReference frame accelerating or rotating, such as on an object at rest on Earth or in motion relative to the Earth
True forceForce that exists when an observation is made from an inertial reference frame-Gravitational force, pressure-gradient force, viscous force
Apparent (inertial) forceFictitious force that appears to exist when an observation is made from a noninertial reference frame but is an acceleration from an inertial reference frame-Apparent centrifugal force, apparent Coriolis force
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Newton’s Second Law of MotionNewton’s second law of motion
Inertial acceleration (4.12)Momentum equation in inertial reference frame
Expand left side of momentum equation (4.15,6)
F =Ma
ai =1
MaF∑
Absolute velocity (4.13)
vA =v+Ω×Re
ai =dvAdt
+Ω×vA =dvdt
+Ω×dRedt
+Ω×v+Ω× Ω×Re( )
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Angular Velocity
Angular velocity magnitude
Ω=jϕΩcosϕ+krΩsinϕ
Fig. 4.3
kr ΩsinϕϕRe
Ω
Re
ΩjϕΩcosϕϕ
Ω =2π rad
86,164 s=7.292×10−5 rad
s
Angular velocity vector (4.14)
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Inertial AccelerationInertial acceleration (4.16)
Total derivative of radius of the Earth vector (4.17)
Vector giving radius of Earth (4.14)
--> Inertial acceleration (4.18)
ai =dvdt
+Ω×dRedt
+Ω×v+Ω× Ω×Re( )
Re =krRe
dRedt
=Redkrdt
=iλu +jϕv≈v
dkrdt
=iλuRe
+jϕvRe
ai =dvdt
+2Ω×v+Ω× Ω×Re( )=al +ac +ar
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Inertial AccelerationLocal, Coriolis, Earth’s centripetal acceleration vectors (4.19)
Treat Coriolis, centripetal accelerations as apparent forces
Expand both sides of momentum equation (4.12) (4.20)
al +ac +ar =1
MaFg
* +Fp+Fv( )
ac=Fc Ma ar =−Fr Ma
al =dvdt
ac =2Ω×v ar =Ω× Ω×Re( )
Momentum equation from Earth’s reference frame (4.21)
al =1
MaFr −Fc +Fg
* +Fp +Fv( )
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Local AccelerationExpand local acceleration (4.22)
al =dvdt
=∂v∂t
+ v•∇( )v
dvdt
=d iu+jv+kw( )
dt=i
dudt
+jdvdt
+kdwdt
Expand left side in Cartesian/altitude coordinates (4.23)
Expand further in terms of local derivative (4.24)
+k∂w∂t
+u∂w∂x
+v∂w∂y
+w∂w∂z
⎛
⎝ ⎜
⎞
⎠ ⎟
=i∂u∂t
+u∂u∂x
+v∂u∂y
+w∂u∂z
⎛
⎝ ⎜
⎞
⎠ ⎟ +j
∂v∂t
+u∂v∂x
+v∂v∂y
+w∂v∂z
⎛
⎝ ⎜
⎞
⎠ ⎟
dvdt
=∂v∂t
+ v•∇( )v=∂∂t
+u∂∂x
+v∂∂y
+w∂∂z
⎛
⎝ ⎜
⎞
⎠ ⎟ iu+jv+kw( )
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Local AccelerationExpand left side in spherical-altitude coordinates (4.25)
Total derivative in spherical-altitude coordinates (4.26)
ddt
=∂∂t
+u1
Recosϕ∂
∂λe+v
1Re
∂∂ϕ
+w∂∂z
diλdt
=jϕutanϕ
Re−kr
uRe
djϕdt
=−iλutanϕ
Re−kr
vRe
dkrdt
=iλuRe
+jϕvRe
dvdt
=d iλu+jϕv+krw( )
dt= iλ
dudt
+udiλdt
⎛ ⎝ ⎜
⎞ ⎠ ⎟ + jϕ
dvdt
+vdjϕdt
⎛
⎝ ⎜
⎞
⎠ ⎟ + kr
dwdt
+wdkrdt
⎛ ⎝ ⎜ ⎞
⎠ ⎟
Total derivative of unit vectors (4.28)
Substitute into (4.25) (4.29)
dvdt
=iλdudt
−uvtanϕ
Re+
uwRe
⎛
⎝ ⎜
⎞
⎠ ⎟ +jϕ
dvdt
+u2 tanϕ
Re+
vwRe
⎛
⎝ ⎜
⎞
⎠ ⎟ +kr
dwdt
−u2
Re−
v2
Re
⎛
⎝ ⎜
⎞
⎠ ⎟
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Example 4.2u = 20 m s-1 Δx = 500 km Re = 6371 kmv = 10 m s-1 Δy = 500 km = 45 oNw = 0.01 m s-1 Δz = 10 km -->
Simplify local acceleration (4.30)
dudt
≈8×10−4 uvtanϕRe
≈3.1×10−5 uwRe
≈3.1×10−8
dvdt
≈2×10−4 u2 tanϕRe
≈6.3×10−5 vwRe
≈1.6×10−8
dwdt
≈1×10−8 u2
Re≈6.3×10−5 v2
Re≈1.6×10−5
dvdt
=iλdudt
−uvtanϕ
Re
⎛
⎝ ⎜
⎞
⎠ ⎟ +jϕ
dvdt
+u2 tanϕ
Re
⎛
⎝ ⎜
⎞
⎠ ⎟ +kr
dwdt
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Local Acceleration
Local acceleration in spherical-altitude coordinates (4.31)
+jϕ∂v∂t
+u
Recosϕ∂v∂λe
+vRe
∂v∂ϕ
+w∂v∂z
+u2 tanϕ
Re
⎛
⎝ ⎜
⎞
⎠ ⎟
+kr∂w∂t
+u
Recosϕ∂w∂λe
+vRe
∂w∂ϕ
+w∂w∂z
⎛
⎝ ⎜
⎞
⎠ ⎟
ddt
=∂∂t
+u1
Recosϕ∂
∂λe+v
1Re
∂∂ϕ
+w∂∂z
dvdt
=iλdudt
−uvtanϕ
Re
⎛
⎝ ⎜
⎞
⎠ ⎟ +jϕ
dvdt
+u2 tanϕ
Re
⎛
⎝ ⎜
⎞
⎠ ⎟ +kr
dwdt
dvdt
=iλ∂u∂t
+u
Recosϕ∂u∂λe
+vRe
∂u∂ϕ
+w∂u∂z
−uvtanϕ
Re
⎛
⎝ ⎜
⎞
⎠ ⎟
Local acceleration in Cartesian-altitude coordinates (4.30)
Total derivative in spherical-altitude coordinates (4.26)
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Equator
South Pole
North Pole
Apparent Coriolis Force
EastWest
Direction of Earth’srotation
A
B B’ C
D A’
E
F F’ G
E’H
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Apparent Coriolis ForceApparent Coriolis force per unit mass (4.32)
Consider only zonal (west-east) wind (4.33)
al =dvdt
=−FcMa
=−jϕ2Ωusinϕ+kr2Ωucosϕ
Equate local acceleration (4.21) with Coriolis force
2Ωukr2Ωucosϕ-jϕ2Ωusinϕϕ
Re
Ω
Re+jϕ2Ωusinϕ −kr2Ωucosϕ
FcMa
=2Ω×v=2Ω
iλ jϕ kr0 cosϕ sinϕ
u v w
=iλ2Ω wcosϕ−vsinϕ( )
FcMa
=2Ω×v
=jϕ2Ωusinϕ−kr2Ωucosϕ
Fig. 4.5
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Apparent Coriolis Force
Coriolis parameter (4.35)
Rewrite (4.34) (4.36)
Eliminate vertical velocity termEliminate k term
--> Apparent Coriolis force per unit mass (4.34)FcMa
=2Ω×v≈−iλ2Ωvsinϕ+jϕ2Ωusinϕ
f =2Ωsinϕ
FcMa
≈−iλ fv+jϕ fu= f
iλ jϕ kr0 0 1
u v 0
= fkr ×vh
FcMa
= f vh = f u2+v2Magnitude
vh =10ms
at North Pole→FcMa
=0.001454m2
sExample
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Gravitational ForceTrue gravitational force vector (4.37)
Fg*
Ma=−kr
*g*
Newton’s law of universal gravitation (4.38)
True gravitational force vector for Earth (4.39)
Equate (4.37) and (4.39) (4.40)
Fg*
Ma=−kr
* GMeRe
2
Fg*
Ma=g* =
GMeRe
2
Me=5.98 x 1024 kg, Re=6370 km -->g*=9.833 m s-2
F12,g =−r21GM1M2
r213 r21=kr
*r21=kr*Re
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Apparent Centrifugal Force
Apparent centrifugal force per unit mass (4.41)
where
FrMa
=−ar =−Ω× Ω×Re( ) =−Ω
iλ* jϕ
* kr*
0 cosϕ sinϕ
Recosϕ 0 0
=−jϕ* ReΩ
2cosϕsinϕ+kr*ReΩ
2cos2ϕ
To observer fixed in space, objects moving with the surface of a rotating Earth exhibit an inward centripetal acceleration. An observer on the surface of the Earth feels an outward apparent centrifugal force.
Ω×Re =i*ReΩcosϕΩ=jϕ* Ωcosϕ+kr
*Ωsinϕ
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Effective GravityAdd gravitational and apparent centrifugal force vectors (4.44)
Effective gravitational acceleration (4.45)
FgMa
=Fg*
Ma+
FrMa
=−jϕ* ReΩ
2cosϕsinϕ+kr* ReΩ
2cos2ϕ−g*( )=−krg
g =ReΩ
2cosϕsinϕ( )2
+ g* −ReΩ2cos2ϕ( )
2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
12
ϕRe
Ω
ReRe cos ϕ*Fg
Frkrkr*
Fg
Fig. 4.6
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Examplesg = 9.799 m s-2 at Equator at sea level
= 9.833 m s-2 at North Pole at sea level--> 0.34% diff. in gravity between Equator and Pole
0.33% diff. (21 km) in Earth radius between Equator and Pole--> Apparent centrifugal force has caused Earth’s Equatorial bulge
g = 9.8060 m s-2 averaged over Earth’s topographical surface, which averages 231.4 m above sea level
Example 4.6 g = 9.497 m s-2 100 km above Equator(3.1% lower than
surface value)--> variation of gravity with altitude much greater than variation of gravity with latitude
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Geopotential
Magnitude of geopotential (4.46)
Geopotential height (4.47)
Work done against gravity to raise a unit mass of air from sea level to a given altitude. It equals the gravitational potential energy of air per unit mass.
Z=Φ z( )g0
Gradient of geopotential (4.48)
FgMa
=−krg =−∇Φ
Effective gravitational force vector per unit mass (4.49)
Φ z( ) = g z( )d0
z
∫ z ≈g0z
∇Φ z( ) =∇Φ =kr∂Φ z( )
∂z≈krg0 =krg
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Pressure-Gradient Force
Sum forces
Forces acting on box (4.50)
Pressure-gradient force per unit mass (4.51)
Fp,rFp,l
Δx
Δy
ΔzFp,r =− pc +
∂p∂x
Δx2
⎛ ⎝ ⎜
⎞ ⎠ ⎟ ΔyΔz
Fp,l = pc −∂p∂x
Δx2
⎛ ⎝ ⎜
⎞ ⎠ ⎟ ΔyΔz
Fp,x =−∂pa∂x
ΔxΔyΔz Ma =ρaΔxΔyΔz
Fp,xMa
=−1ρa
∂pa∂x
Mass of air parcel
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Pressure-Gradient Force ExampleL1012 hPa1008 hPa
100 kmH
1ρa
∂pa∂x
≈1
1.2 kg m−31012−1008 hPa
105 m
⎛
⎝ ⎜
⎞
⎠ ⎟
100 kg m−1 s−2
hPa=0.0033 m s−2
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Pressure-Gradient ForceCartesian-altitude coordinates (4.52)
FpMa
=−1
ρa∇pa =−
1ρa
i∂pa∂x
+j∂pa∂y
+k∂pa∂z
⎛
⎝ ⎜
⎞
⎠ ⎟
FpMa
=−1
ρa∇pa =−
1ρa
iλ1
Recosϕ∂pa∂λe
+jϕ1Re
∂pa∂ϕ
+kr∂pa∂z
⎛
⎝ ⎜
⎞
⎠ ⎟
Spherical-altitude coordinates (4.53)
Example 4.8 z = 0 m --> pa = 1013 hPaz = 100 m --> pa = 1000
hPaa = 1.2 kg m-3
--> PGF in the vertical 3000 times that in the horizontal:
1ρa
∂pa∂z
≈1
1.2 kg m−31013−1000 hPa
100 m⎛ ⎝ ⎜ ⎞
⎠ ⎟ 100 kg m-1 s−2
hPa=10.8
m
s2
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ViscosityViscosity in liquids
Internal friction when molecules collide and briefly bond. Viscosity decreases with increasing temperature.
Viscosity in gasesTransfer of momentum between colliding molecules. Viscosity increases with increasing temperature.
Dynamic viscosity of air (kg m-1 s-1) (4.54)
ηa =5
16Ada2
maR*Tπ
≈1.8325×10−5 416.16T +120
⎛ ⎝ ⎜ ⎞
⎠ ⎟
T296.16
⎛ ⎝ ⎜ ⎞
⎠ ⎟ 1.5
Kinematic viscosity of air (m2 s-1) (4.55)
νa =ηaρa
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ViscosityWind shear
Change of wind speed with height
Shearing stressViscous force per unit area resulting from shear
Shearing stress in the x-z plane (N m-2) (4.56)
Force per unit area in the x-direction acting on the x-y plane (normal to the z-direction)
τzx=ηa∂u∂z
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Viscous ForceShearing stress in the x-direction
Net viscous force on parcel in x-direction (4.58)
Viscous force after substituting shearing stress (4.59)
Δx
Δy
Δzτzx, top
τzx,botτzx
Fv,zxMa
= τzx,top−τzx,bot( )ΔxΔy
ρaΔxΔyΔz=
1ρa
∂τzx∂z
Fv,zxMa
=1ρa
∂∂z
ηa∂u∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟ ≈
ηaρa
∂2u
∂z2
τzx,top=τzx+∂τzx∂z
Δz2
τzx,bot=τzx−∂τzx∂z
Δz2
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Viscous Force
Viscous force as function of wind shear (4.59)
uz
Net viscous forceτzx,mid
uzτzx,botτzx,top
Nonetviscousforce xxτzx,midτzx,botτzx,top
Fig. 4.10
Fv,zxMa
=1ρa
∂∂z
ηa∂u∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟ ≈
ηaρa
∂2u
∂z2
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Three-Dimensional Viscous ForceExpand (4.58) (4.60)
FvMa
=ηaρa
∇2v=νa∇2v
∇2v= ∇ •∇( )v=i∂2u
∂x2 +∂2u
∂y2 +∂2u
∂z2⎛
⎝ ⎜
⎞
⎠ ⎟
+j∂2v
∂x2 +∂2v
∂y2 +∂2v
∂z2⎛
⎝ ⎜
⎞
⎠ ⎟ +k
∂2w
∂x2 +∂2w
∂y2 +∂2w
∂z2⎛
⎝ ⎜
⎞
⎠ ⎟
Gradient term (4.61)
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Viscous Force Example
--> Viscous force per unit mass aloft is small
Fv,zxMa
≈ηaρa
1z3 −z1( ) 2
u3 −u2z3 −z2
−u2 −u1z2 −z1
⎛
⎝ ⎜
⎞
⎠ ⎟ =5.17×10−10 m
s2
Example 4.9 z1 = 1 km u1 = 10 m s-1 z2 = 1.25 km u2 = 14 m
s-1 z3 = 1.5 km u3 = 20 m
s-1 T = 280 K a = 1.085
kg m-3
--> a = 0.001753 kg m-1 s-2
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Viscous Force Example
--> Viscous force per unit mass at surface is comparable with horizontal pressure-gradient force per unit mass
Fv,zxMa
≈ηaρa
1z3 −z1( ) 2
u3 −u2z3 −z2
−u2 −u1z2 −z1
⎛
⎝ ⎜
⎞
⎠ ⎟ =1.17×10−3 m
s2
Example 4.10 z1 = 0 m u1 = 0 m s-1 z2 = 0.05 m u2 = 0.4 m
s-1 z3 = 0.1 m u3 = 1 m s-
1 T = 288 K a = 1.225
kg m-3
--> a = 0.001792 kg m-1 s-2
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Turbulent Flux DivergenceLocal acceleration (4.22)
al =dvdt
=∂v∂t
+ v•∇( )v
∂ρa∂t
+ρa ∇•v( )+ v•∇( )ρa =0
ρaal =ρa∂v∂t
+ v•∇( )v⎡ ⎣ ⎢
⎤ ⎦ ⎥ +v
∂ρa∂t
+∇ • vρa( )⎡ ⎣ ⎢
⎤ ⎦ ⎥
ρa ≈ρ a v=v + ′ v
a l =∂v ∂t
+ v •∇( )v ′ a l =FtMa
=1ρ a
ρ a ′ v •∇( ) ′ v + ′ v ∇ • ′ v ρ a( )[ ]
Continuity equation for air (3.20)
Combine (4.62)
Decompose variables
Reynolds average (4.62) (4.65)
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Turbulent Flux Divergence
FtMa
=i1
ρa
∂ ρa ′ u ′ u ( )
∂x+
∂ ρa ′ v ′ u ( )
∂y+
∂ ρa ′ w ′ u ( )
∂z
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
+j1ρa
∂ ρa ′ u ′ v ( )
∂x+
∂ ρa ′ v ′ v ( )
∂y+
∂ ρa ′ w ′ v ( )
∂z
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
+k1ρa
∂ ρa ′ u ′ w ( )
∂x+
∂ ρa ′ v ′ w ( )
∂y+
∂ ρa ′ w ′ w ( )
∂z
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
Expand turbulent flux divergence (4.66)
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Diffusion Coefficients for Momentum
Vertical kinematic turbulent fluxes from K-theory (4.67)
′ w ′ u =−Km,zx∂u ∂z
′ w ′ v =−Km,zy∂v ∂z
FtMa
=−i1
ρa
∂∂x
ρaKm,xx∂u∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟ +
∂∂y
ρaKm,yx∂u∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zx∂u∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
−j1ρa
∂∂x
ρaKm,xy∂v∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟ +
∂∂y
ρaKm,yy∂v∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zy∂v∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
−k1ρa
∂∂x
ρaKm,xz∂w∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟ +
∂∂y
ρaKm,yz∂w∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zz∂w∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
Substitute fluxes into turbulent flux divergence (4.68)
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Diffusion Coefficients for Momentum
Turbulent flux divergence in vector/tensor notation (4.70)
FtMa
=−1
ρa∇ •ρaKm∇( )v
Km=
Km,xx 0 0
0 Km,yx 0
0 0 Km,zx
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ for u
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Diffusion Coefficient Examples
Ft,zxMa
=1ρa
∂∂z
ρaKm,zx∂u∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟ ≈
Km,zxz3 −z1( ) 2
u3−u2z3−z2
−u2−u1z2−z1
⎛
⎝ ⎜
⎞
⎠ ⎟ =0.02
m
s2
Ft,yxMa
=1ρa
∂∂y
ρaKm,yx∂u∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ ≈
Km,yxy3−y1( ) 2
u3 −u2y3 −y2
−u2 −u1y2 −y1
⎛
⎝ ⎜
⎞
⎠ ⎟ =−0.0004
m
s2
Example 4.11 Vertical diffusion in middle of boundary layer z1 = 300 m u1 = 10 m s-1
z2 = 350 m u2 = 12 m s-1
z3 = 400 m u3 = 15 m s-1
Km = 50 m2 s-1 --> Example 4.12 Horizontal diffusion y1 = 0 m u1 = 10 m s-1
y2 = 500 m u2 = 9 m s-1
y3 = 1000 m u3 = 7 m s-1
Km = 100 m2 s-1
-->
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Momentum Equation TermsTerm Acceleration or Force /
Mass ExpressionHorizontal
Accel.
(m s-2)
VerticalAccel.
(m s-2)Localacceleration
a l =d vdt
=∂ v∂t
+ v • ∇( ) v 10-4 10-7-1
Coriolis forceperunitmass
FcMa
= fk ×v 10-3 0
Effectivegravitationalforce per unitmass
FgMa
=Fg*
Ma+
FrMa
=−∇Φ0 10
Pressuregradient forceperunitmass
FpMa
=−1a
∇pa10-3 10
Viscous forceperunitmass
FvMa
=aa
∇2v 10-12-10-3 10-15-10-5
Turbulent fluxdivergence ofmomentum
FtMa
=−1a
∇• aK m∇( )v 0-0.005 0-1
Table 4.1
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Momentum Equation
Momentum equation in three dimensions (4.71)
dvdt
=−fk×v−∇Φ−1ρa
∇pa +ηaρa
∇2v+1ρa
∇ •ρaKm∇( )v
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Momentum Equation in Cartesian-Altitude Coordinates
U-direction (4.73)dudt
=∂u∂t
+u∂u∂x
+v∂u∂y
+w∂u∂z
= fv−1ρa
∂pa∂x
+1
ρa
∂∂x
ρaKm,xx∂u∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟ +
∂∂y
ρaKm,yx∂u∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zx∂u∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
dvdt
=∂v∂t
+u∂v∂x
+v∂v∂y
+w∂v∂z
=−fu−1ρa
∂pa∂y
+1
ρa
∂∂x
ρaKm,xy∂v∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟ +
∂∂y
ρaKm,yy∂v∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zy∂v∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
dwdt
=∂w∂t
+u∂w∂x
+v∂w∂y
+w∂w∂z
=−g −1ρa
∂pa∂z
+1
ρa
∂∂x
ρaKm,xz∂w∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟ +
∂∂y
ρaKm,yz∂w∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zz∂w∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
V-direction (4.74)
W-direction (4.75)
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Momentum Equation in Spherical-Altitude Coordinates
U-direction
V-direction
W-direction (4.78)
+1
ρa
1
Re2cosϕ
∂∂λe
ρaKm,xzcosϕ
∂w∂λe
⎛
⎝ ⎜
⎞
⎠ ⎟ +
1
Re2
∂∂ϕ
ρaKm,yz∂w∂ϕ
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zz∂w∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
∂u∂t
+u
Recosϕ∂u∂λe
+vRe
∂u∂ϕ
+w∂u∂z
=uvtanϕ
Re+fv−
1ρaRecosϕ
∂pa∂λe
+1
ρa
1
Re2cosϕ
∂∂λe
ρaKm,xxcosϕ
∂u∂λe
⎛
⎝ ⎜
⎞
⎠ ⎟ +
1
Re2
∂∂ϕ
ρaKm,yx∂u∂ϕ
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zx∂u∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
∂v∂t
+u
Recosϕ∂v∂λe
+vRe
∂v∂ϕ
+w∂v∂z
=−u2 tanϕ
Re− fu−
1ρaRe
∂pa∂ϕ
+1
ρa
1
Re2cosϕ
∂∂λe
ρaKm,xycosϕ
∂v∂λe
⎛
⎝ ⎜
⎞
⎠ ⎟ +
1
Re2
∂∂ϕ
ρaKm,yy∂v∂ϕ
⎛
⎝ ⎜
⎞
⎠ ⎟ +
∂∂z
ρaKm,zy∂v∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
∂w∂t
+u
Recosϕ∂w∂λe
+vRe
∂w∂ϕ
+w∂w∂z
=−g−1ρa
∂pa∂z
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Scaling ParametersEkman, Rossby, Froude numbers (4.72)
Ek =νau/ x2
ufRo=
u2 / xuf
Fr2 =w2 / z
g
Example 4.13 a = 10-6 m2 s-1 u = 10 m s-1 x = 106 m w
= 0.01 m s-1 z = 104 m f
= 10-4 s-1 --> Ek = 10-14
--> Ro = 0.1--> Fr = 0.003
Viscous accelerations negligible over large scalesCoriolis more important than local horizontal accelerationsGravity more important than vertical inertial accelerations
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Geostrophic WindGeostrophic Wind (4.79)
Elim. all but pressure-gradient, Coriolis terms in momentum eq.
vg =1fρa
∂pa∂x
ug =−1
fρa
∂pa∂y
vg =iug +jvg =1
fρa−i
∂pa∂y
+j∂pa∂x
⎛
⎝ ⎜
⎞
⎠ ⎟ =
1fρa
i j k
0 0 1∂pa∂x
∂pa∂y
0
=1
fρak×∇zpa
∇z = i∂∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟
z+ j
∂∂y
⎛
⎝ ⎜
⎞
⎠ ⎟
z=i
∂∂x
+j∂∂y
Example 4.14 = 30o a = 0.00076 g cm-3
∂pa/∂y = 4 hPa per 150 km--> f = 7.292x10-5 s-1 --> ug = 48.1 m s-1
Geostrophic Wind in cross-product notation (4.80)
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Surface WindsFig. 4.11. Force and wind vectors aloft and at surface in Northern Hemisphere.
−fv=−1ρa
∂pa∂x
+1ρa
∂∂z
ρaKm,zx∂u∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
fu=−1
ρa
∂pa∂y
+1ρa
∂∂z
ρaKm,zy∂v∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
HL Surface
HL Fp
Aloft
Fp
Fc
Fc
FtFt+Fc
v
vHorizontal equation of motion near the surface (4.82)
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Boundary-Layer Winds
Cloud layerEntrainment zoneInversion layerFree troposphere
Neutralconvectivemixed layerSurface layer
Subcloud layer
Daytime mean wind speed
Boundary layerAltitudeEntrainment zoneInversion layerFree troposphere
Surface layerNighttime mean wind speed
Residual layerStableboundarylayerNocturnal jet
Boundary layerAltitude
Fig. 4.12
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Morning/Afternoon Observed Winds at Riverside
-15 -10 -5 0 5 10 15
700
750
800
850
900
950
1000
Wind speed (m s-1)
Pressure (hPa)
3:30 a.m.
u v
-15 -10 -5 0 5 10 15
700
750
800
850
900
950
1000
Wind speed (m s-1)
Pressure (hPa)
u v
3:30 p.m.
Fig. 4.13
Pre
ssur
e (h
Pa)
Pre
ssur
e (h
Pa)
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Gradient WindCartesian to cylindrical coordinate conversions (4.83)
x=Rc cosθ y=Rcsinθ
Rc2 =x2 +y2 θ =tan−1 y
x⎛ ⎝ ⎜ ⎞
⎠ ⎟
Rc =iRRc
uR =dRcdt
vθ =Rcdθdt
ijRc
jθ iR
ORcθ PvθuRaR
Fig. 4.14
Radial vector (4.86)
Radial and tangential scalar velocities (4.86)
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Gradient Wind
Horizontal momentum equation without turbulence (4.91)
L FcFRFpvθ
834hPa830hPa HFc FRFpvθ
834hPa830hPa
Fig. 4.15
duRdt
= fvθ−1ρa
∂pa∂Rc
+vθ2
Rc
vθ =−Rc f2
±Rc2
f2+41
Rcρa
∂pa∂Rc
Remove local acceleration, solve (4.92)
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Gradient Wind ExampleGradient wind speed (4.92)
Example 4.15 Low pressure near center of hurricanepa/Rc = 45 hPa per 100 km Rc = 70 km = 15
pa = 850 hPa a = 1.06 kg m-3 --> vθ = 52 m s-1 --> vg= 1123 m s-1
vθ =−Rc f2
±Rc2
f2+41
Rcρa
∂pa∂Rc
High-pressure centerpa/Rc = -0.1 hPa per 100 km --> vθ = -1.7 m s-1 --> vg = 2.5 m s-1
--> pressure gradient and gradient wind lower around high-pressure center than low-pressure center.
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Surface Winds Around Lows/Highs
Momentum equations for surface winds (4.93)
L FcFRFpvθ1000hPa996hPa
Ft HFcFRFpvθ
1012hPa1016hPaFt
duRdt
= fvθ−1ρa
∂pa∂Rc
+vθ2
Rc+
1ρa
∂ ρa ′ w ′ u R( )
∂z
dvθdt
=−fuR −uRvθRc
+1
ρa
∂ ρa ′ w ′ v θ( )
∂z
Fig. 4.16
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Atmospheric Waves
Displacement and amplitude (4.98)D x,t( ) =Aw sin ˜ k x−ναt( )
˜ k =2π
λα,x
˜ l =2π
λα,y˜ m =
2πλα,z
Wavenumber and wavelength (4.95)
0 2 4 6 8 10-2
-1
0
1
2
Displacement (m)
x
Wavelength
(m)
Amplitude
Fig. 4.17
Dis
plac
emen
t (m
)
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Atmospheric Waves
Frequency of oscillation (dispersion relationship) (4.97)Phase speed c = speed at which all components of the individual wave travel along the direction of propagation.
˜ K = ˜ k 2 +˜ l 2 + ˜ m 2
να =cα˜ k 2+˜ l 2 + ˜ m 2 =cα
˜ K
Wavenumber vector (4.94)
Superposition principleDisplacement of a medium due to a group of waves of different wavelength equals the sum of displacements due to each individual wave in the group.
EnvelopeShape of the sum of the waves (shape of the group)
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Group VelocityGroup velocity vector and group speed (4.99)
Velocity of envelope of group
cg,x =∂να∂˜ k
=cα˜ k ˜ K
+ ˜ K ∂cα∂˜ k
cg,y =∂να∂˜ l
=cα˜ l ˜ K
+ ˜ K ∂cα∂˜ l
cg,z =∂να∂ ˜ m
=cα˜ m ˜ K
+ ˜ K ∂cα∂ ˜ m
cg=icg,x +jcg,y +kcg,z cg = cg,x2 +cg,y
2 +cg,z2
Group scalar speeds (4.101)
να =cα˜ k 2+˜ l 2 + ˜ m 2 =cα
˜ K where
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Nondispersive/Dispersive MediaNondispersive medium (4.103)
Phase speed independent of group speed
∂cα∂˜ k
=∂cα∂˜ l
=∂cα∂ ˜ m
=0
cg,x =cα˜ k ˜ K
cg,y =cα˜ l ˜ K
cg,z =cα˜ m ˜ K
∂cα∂˜ k
≠0∂cα∂˜ l
≠0∂cα∂ ˜ m
≠0
Dispersive mediumPhase speed dependent on group speed
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Nondispersive/Dispersive MediaSound waves Water waves
Fig. 4.18
0 0.5 1 1.5 2
Displacement
Time 1
Time 2
Nondispersive wave
x(m)
να =cs˜ k
cg =cg,x =∂να∂˜ k
=cs =cα
Dis
plac
emen
t (m
)
0 0.5 1 1.5 2
Displacement
Time 1
Time 2
Dispersive wave
x(m)
να = g˜ k
cg =cg,x =∂να∂˜ k
=12
g˜ k
=12
cα
Dis
plac
emen
t (m
)
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Acoustic (Sound) WavesOccur when a vibration causes alternating adiabatic compression and expansion of a compressible fluid, such as air. During compression/expansion, air pressure oscillates, causing acceleration to oscillate along the direction of propagation of the wave.
dudt
=−1ρa
∂pa∂x
dρadt
=−ρa∂u∂x
1θv
dθvdt
=dlnθv
dt=0
U-momentum equation (4.105)
Continuity equation for air (4.106)
Thermodynamic energy equation (4.107)
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Acoustic (Sound) Waves
dρadt
=ρaγ
dlnpadt
γ =1
1−κ=
cp,dcv,d
≈1.4
Substitute θv =Tv 1000pa( )κ and pa =ρa ′ R Tv into (4.107)
--> Revised thermodynamic energy equation (4.108)
Substitute (4.108) into continuity equation (4.106) (4.109)dlnpa
dt=
1pa
dpadt
=−γ∂u∂x
1θv
dθvdt
=dlnθv
dt=0
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Acoustic Wave Equation
Speed of sound under adiabatic conditions (4.111)
d2 ′ p adt2
=∂∂t
+u ∂∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2
′ p a =cs2 ∂2 ′ p a
∂x2
cs =± γ ′ R T v
′ p a = ′ p a,0sin ˜ k x−ναt( )
να = u ±cs( )˜ k
cg =cg,x =cα =u ±cs
Take time derivative of (4.109) and combine with momentum equation (4.106) --> acoustic wave equation (4.110)
Solution to wave equation (4.112)
Dispersion relationship for acoustic waves (4.113)
Group speed equals phase speed --> nondispersive (4.114)
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Acoustic-Gravity WavesGravity waves
When the atmosphere is stably stratified and a parcel of air is displaced vertically, buoyancy restores the parcel to its equilibrium position in an oscillatory manner.
Acoustic-gravity wave dispersion relationship found as follows:
dρadt
=−ρa∂u∂x
+∂w∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟
dρadt
=ρaγ
dlnpadt
Momentum equations retaining gravity (4.115)
Continuity equation
Thermodynamic energy equation from acoustic case
dudt
=−1ρa
∂pa∂x
dwdt
=−1ρa
∂pa∂z
−g
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Acoustic-Gravity Waves
Nbv2
να −u ̃ k ( )2
˜ k 2+να −u ̃ k ( )
2
cs2 = ˜ m 2+˜ k 2+
νc2
cs2
νc =cs2H
Acoustic-gravity wave dispersion relationship (4.116)
Acoustic cutoff frequency (4.117)
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Acoustic-Gravity Waves
να =
u ̃ k +Nbv˜ k
˜ k 2+ ˜ m 2 +νc2 cs
2( )
12 να2 <<νc
2 low−frequency
gravitywaves⎧ ⎨ ⎩
u ̃ k + cs2˜ k 2+cs
2˜ m 2+νc2
( )12
να2 >>Nbv
2 high−frequency
acousticwaves⎧ ⎨ ⎩
u ̃ k +Nbv
˜ k
˜ k 2+ ˜ m 2( )12
˜ k → ∞ mountainlee waves{
u +cs( )˜ k να2 <<νc
2, ˜ m 2 =0, ˜ k → 0 or
να2 >>Nbv
2 , ˜ m 2 =0, ˜ k → ∞
⎧ ⎨ ⎪
⎩ ⎪
Lamb
waves⎧ ⎨ ⎩
⎧
⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎩
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 5 10-5 0.0001 0.00015 0.0002
-(s
-1 )
k
2=0~m 2>0~m
2<0~m
2=0~m
νc
Nbv
Lamb waves
High-frequency internalacoustic waves
Low-frequency internalgravity waves
~
2>0~m
(m-1)
uk~
c ks
~
~
Fig. 4.19
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Inertial Oscillation
Horizontal momentum equations with Coriolis (4.121)
u y0+Δy( )−ug y0( ) ≈ fΔy
When a parcel of air moving from west to east is perturbed in the south-north direction, the Coriolis force propels the parcel toward its original latitude in an inertially stable atmosphere and away from its original latitude in an inertially unstable atmosphere. In the former case, the parcel subsequently oscillates about its initial latitude in an inertial oscillation.
dudt
= fv=fdydt
dvdt
= f ug−u( )
Integrate u-equation between y0 and y0+Δy (4.123)
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Inertial OscillationsTaylor-series expansion of geostrophic wind (4.124)
ug y0 +Δy( ) ≈ug y0( )+∂ug∂y
Δy
ug y0 +Δy( ) −u y0+Δy( )≈− f −∂ug∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ Δy
dvdt
=−f f −∂ug∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ Δy
f −∂ug∂y
<0 inertially unstable
=0 inertially neutral
>0 inertially stable
⎧
⎨ ⎪
⎩ ⎪
Substitute (4.124) into (4.123) (4.125)
Substitute (4.125) into v-momentum equation (4.126)
Inertial stability criteria in Northern Hemisphere (4.127)
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Inertial Lamb and Gravity WavesInertial Lamb waves (4.128)
να2 = f2 +cs
2˜ k 2
να2 = f2 +
Nbv2 ˜ k 2
˜ k 2 + ˜ m 2+νc2 cs
2
λR =ghef
he=
cs2
ginertia Lamb waves
Nbv2 g
˜ m 2 +νc2 cs
2 inertia gravity waves
⎧
⎨
⎪ ⎪
⎩
⎪ ⎪
Inertial gravity waves (4.129)
Rossby radius of deformation (4.130)L>λR --> velocity field adjusts to pressure field
Equivalent depth (4.131)
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Geostrophic Adjustment
H L
Geostrophicwind
570 hPa566 hPa
Eastxy
PressuregradientforceCoriolisforce
150 km
Fig. 4.20
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VorticityRelative vorticity (4.132)
ζr =∇ ×v=
i j k∂∂x
∂∂y
∂∂z
u v w
=∂w∂y
−∂v∂z
⎛
⎝ ⎜
⎞
⎠ ⎟ i−
∂w∂x
−∂u∂z
⎛ ⎝ ⎜
⎞ ⎠ ⎟ j+
∂v∂x
−∂u∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ k
ζr,z =∂v∂x
−∂u∂y
ζa,z = f +ζr,z
Pv =f +ζr,z
Δzt=
f +∂v∂x
−∂u∂y
Δzt=constant
Vertical component of relative vorticity
Absolute vorticity
Potential vorticity (4.133)
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Rossby WavesHorizontal momentum equations (4.134)
dudt
= fv−1
ρa
∂pa∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟
z
dvdt
=−fu−1ρa
∂pa∂y
⎛
⎝ ⎜
⎞
⎠ ⎟
z
f = f0+β y−y0( )
β =∂f∂y
=2Ω∂ϕ∂y
cosϕ≈2ΩRe
cosϕ
∂pa∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟
z=ρa
∂Φ∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟
p
∂pa∂y
⎛
⎝ ⎜
⎞
⎠ ⎟
z=ρa
∂Φ∂y
⎛
⎝ ⎜
⎞
⎠ ⎟
p
u =ug+ua Φ =Φg+Φa
Midlatitude beta-plane approximations (4.136)
Geopotential gradients on surfaces of constant pressure (4.138)
Separate variables into geostrophic/ageostrophic components
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Rossby Waves
Rewrite momentum equations (4.140,1)
d ug +ua( )
dt= f0 +β y−y0( )[ ] vg +va( )−
∂ Φg+Φa( )
∂x
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
p
d vg+va( )
dt=− f0 +β y−y0( )[ ] ug +ua( )−
∂ Φg+Φa( )
∂y
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
p
vg =1f0
∂Φg∂x
⎛
⎝ ⎜
⎞
⎠ ⎟
pug =−
1f0
∂Φg∂y
⎛
⎝ ⎜
⎞
⎠ ⎟
p
Combine geostrophic wind with geopotential gradients (4.142)
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Rossby WavesSubstitute (4.42) into (4.40), (4.41) (4.143,4)
--> quasigeostrophic momentum equations
dugdt
= f0va+β y−y0( )vg −∂Φa∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟
pdvgdt
=−f0ua−β y−y0( )ug −∂Φa∂y
⎛
⎝ ⎜
⎞
⎠ ⎟
p
ddt
∂vg∂x
−∂ug∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ =−f0
∂ua∂x
+∂va∂y
⎛
⎝ ⎜
⎞
⎠ ⎟ −βvg
Subtract ∂/∂y of (4.143) from ∂/∂x of (4.144) (4.145)
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Rossby WavesVertical velocity (4.146)
w =dzdt
=1g
dΦdt
∂ug∂x
+∂vg∂y
=0
∂u∂x
+∂v∂y
+∂w∂z
=0
1g
∂∂z
dΦgdt
⎛
⎝ ⎜
⎞
⎠ ⎟ =−
∂ua∂x
+∂va∂y
⎛
⎝ ⎜
⎞
⎠ ⎟
dΦgdt
=−gΔzt∂ua∂x
+∂va∂y
⎛
⎝ ⎜
⎞
⎠ ⎟
Substitute (4.146), u=ug+ua, v=vg+va and
Continuity equation for incompressible air
to obtain (4.147)
Integrate from surface to mean tropopause height Δzt (4.148)
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Rossby WavesSubstitute (4.148) into (4.145) (4.149)
ddt
ζg −f0
gΔztΦg
⎛
⎝ ⎜
⎞
⎠ ⎟ =−βvg
ζg =∂vg∂x
−∂ug∂y
=1f0
∂2Φg
∂x2 +∂2Φg
∂y2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
p
=∇ p
2Φgf0
∂∂t
+u ∂∂x
⎛ ⎝ ⎜
⎞ ⎠ ⎟ ∇ p
2Φg −f02
gΔztΦg
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ +β
∂Φg∂x
⎛
⎝ ⎜
⎞
⎠ ⎟
p=0
Φg =Φg,0sin ˜ k x+˜ l y−ναt( )
να = u −β
˜ k 2+˜ l 2 +λR−2
⎛
⎝ ⎜
⎞
⎠ ⎟ ̃ k λR =
gΔztf0
Geostrophic potential vorticity (4.150)
Expand (4.149) (4.150) -> quasi-geostrophic potential vorticity equation
Wave solution (4.152)
Dispersion rel. for freely-propagating Rossby waves (4.152)