Presentation Slides for

Chapter 4of

Fundamentals of Atmospheric Modeling 2nd Edition

Mark Z. JacobsonDepartment of Civil & Environmental Engineering

Stanford UniversityStanford, CA 94305-4020jacobson@stanford.edu

March 10, 2005

Spherical Horizontal Coordinates

iλkrRecosϕ

λeReReϕ

jϕ

Fig. 4.1

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

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

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

⎛

⎝ ⎜

⎞

⎠ ⎟

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)

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)

Assume Re constant (4.11)

Spherical Coordinate Conversions

∇ •v=1

Recosϕ∂u∂λe

+1

Recosϕ∂∂ϕ

vcosϕ( )+∂w∂z

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

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( )

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)

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

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( )

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( )

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

⎛

⎝ ⎜

⎞

⎠ ⎟

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

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

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

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)

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)

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)

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

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

-->

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

Momentum Equation

Momentum equation in three dimensions (4.71)

dvdt

=−fk×v−∇Φ−1ρa

∇pa +ηaρa

∇2v+1ρa

∇ •ρaKm∇( )v

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)

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

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

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)

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)

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

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)

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)

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)

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.

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

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

)

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)

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

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

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

)

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)

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

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)

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

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)

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

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)

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)

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)

Geostrophic Adjustment

H L

Geostrophicwind

570 hPa566 hPa

Eastxy

PressuregradientforceCoriolisforce

150 km

Fig. 4.20

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)

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

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)

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)

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)

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)