Reinisch_ASD_85.5151 Chapter 2. Basic Conservation Laws x0x0 y0y0 z0z0 z y x U(x,y,z,t) mass element

Reinisch_ASD_85.515 1

Chapter 2. Basic Conservation Laws

x0

y0

z0

z

y

x

U(x,y,z,t)mass element

x y z

:

Description of mass, momentum, energy in a volume

element x y z at a fixed location.

:

Description of mass, momentum, energy in a volume

element x y z moving with the

Eulerian frame

Lagrangian frame

fluid.


2.1 Total Differentiation(in the Lagrangian frame)

The laws of conservation of mass, momentum, energy

are valid in an inertial frame of reference, i.e., the

Eulerian frame.

The rate of change of a field variable (T, p, ...) in a

Lagrangian parcel is cD

alled the derivative .Dt

Since the coordinates of the parcel are given as

function of time, x=x(t), y=y(t), z=z(t), the total

derivative is truly a function of time only.

The change of field var

total

iable T in the moving parcel

between times t and t+ t is given by the Taylor series


Local Temperature Change and Temperature Advection

T T T TT= ... / , 0 :

t

T T T T

t

, ,

T T T T2.1

t

T

t

T

t x y z t tx y z

DT Dx Dy Dz

Dt x Dt y Dt z Dt

Dx Dy Dzu v w

Dt Dt Dt

DTu v w

Dt x y z

DTT

Dt

U

is the temper advectioature t

nDT

T TDt

U U


Example for Advectionp

East xWest

p(x)

180 km

300

Pa

u=10km/h

The barometer moving with the ship measures

Dpa preasure rate of change of = -0.1 kPa/3h.

DtWhat is the rate of change in an Eulerian parcel?

p Dp 0.30.1 / 3 10

t Dt 180

p0.1 / 6

t

p km kPau kPa h

x h km

kPa

would be measured at fixed point.h


Total Differentiation of a Vector in a Rotating System

Let be an arbitrary vector.

In the inertial frame , , :

1x y zA A A

A

i j k

A i j k

' ' '

In the rotating frame ', ', ' :

' ' ' 2x y zA A A

i j k

A i j k

a

The total derivative in the inertial (absolute) frame

relates to the applied forces (Newton's laws).

D

Dtyx z

DADA DA

Dt Dt Dt

Ai j k


Rotation

'' '' ' 'a a a a

'' '' ' '

But is also given by (2), therefore:

D D ' D ' D '' ' '

Dt Dt Dt DtNotice that

' ' ' ' ' '

is the rate of change of viewed in

yx zx y z

yx zx y z

DADA DAA A A

Dt Dt Dt

DADA DA D DA A A

Dt Dt Dt Dt Dt

A

A i j ki j k

Ai j k i j k

A

a

the rotating system. Also

D '' ', .

Dt

detc

dt

ik i Ω i

i

i’

i’

k

j


Rotating Coordinate System

'' '' ' 'a a a a

' ' 'a

' ' '

a

D D ' D ' D '' ' '

Dt Dt Dt Dtbecomes therefore:

D' ' '

Dt

' ' '

or

D2.2

Dt

yx zx y z

x y z

x y z

DADA DAA A A

Dt Dt Dt

DA A A

DtD

A A ADt

D

Dt

A i j ki j k

A AΩ i Ω j Ω k

AΩ i j k

A AΩ A


2.2 Momentum Equation in a Rotating Coordinate System

Newton's 2nd law of motion (in inertial system):

is the absolute velocity measured in the inertial frame.

(using 2.2)

Here is the change of in the rotating frame:

a a

a

aa

D

Dt

D D

Dt DtD

DtD

Dt

UF

U

r rU Ω r

rr

r .Then 2.5a U U U Ω r


Absolute Derivation of Ua 2.5

Since is the geocenric distance, is the velocity

due to Earth's rotation. The absolute velocity is therefore

equal to the velocity relative to the Earth frame +

the rotation velocity.

a

U U Ω r

r Ω r

a a aa

Using 2.2 :

D, and using 2.5

Dt

.Therefore

D

DtD

DtD

Dt

U UΩ U

U Ω r Ω U Ω r

UΩ U Ω U Ω Ω r

2a aD2 2.7

DT

D

DT

U UΩ U R


2a a

2

2

But from Newton's law

D2

Dt1

* . Therefore

1* 2

But * , so

12 2.8

Momentum equation in the rotating frame.

r

r

r

D

Dt

p

Dp

Dt

Dp

Dt

U UΩ U R F

F g F

Ug F Ω U R

g R g

Ug F Ω U

Acceleration in Rotating Frame


2.3 Using Spherical Coordinates

cos

cos , , 2.9

We define eastward and northward displacements as

cos , . .Then:

; ; ;

R r

D D D Dzu R a v r w

Dt Dt Dt Dt

Dx a D Dy rD r a z

Dx Dy Dzu v w u v w

Dt Dt Dt

U i j k

ik

j

R

r

U

Notice that this Cartesian , , system is not

an inertial one, but is changing when the

parcel moves with .

i j k

U


2.3 cont’d defined in rotating system.

D

Dt

But

u v w

Du Dv Dw D D Du v w

Dt Dt Dt Dt Dt DtD

u v w uDt t x y z x

U i j k

U i j ki j k

i i i i i i

We use Fig. 2.1 to find the magnitude of :

1lim lim lim

cos cos

x

x x x r r

i

ii

The direction of (Fig. 2.2) is - sin cosRx

i

R j k


2.3 cont’d

sin cos

cos

tan . Similarly:

tan2.12

2.13

x r

D u uu

Dt x r rD u v

Dt r rD u v

Dt r r

i j k

i ij k

ji k

ki k

2 2 2

tan

tan2.14

D Du u uv w

Dt Dt r r

Dv u vw Dw u v

Dt r r Dt r

r a z a

Ui

j k


Components of DU/Dt

2

2 2

tan 12 sin 2 cos

tan 12 sin

12 cos

2.19 21

rx

ry

rz

Du vu uw pv w F

Dt r r x

Dv u vw pu F

Dt r r y

Dw u v pu g F

Dt r z


2.4 Scale Analysis (horizontal components)

2

2

0 0 2

tan 12 sin 2 cos

tan 12 sin

/

rx

ry

Du uw vu pv w F

Dt r r x

Dv vw u pu F

Dt r r y

U UW U P Uf U f W

L U a a L H

A B C D E F G

2 2-2 -53-4 -4 -2

26

-4 -3 -6 -8 -5 -3 -12

7 7 6 4

10 10 10 1010 10 1010 10 10 10

10 10 10 1 10

10 10 10 10 10 10 10

10


2.41 Geostrophic Approximation

The DEs for the horizontal velocities are dominated

terms B(Coriolis) and C (pressure gradient).

We approximate these equations:

1 1- = - , = - 2.22

where 2 sin is the Coriolis factor. In v

g g

p pfv fu

x y

f

ector notation:

is the "geostrophic" wind

12.23

g g g

g

u v

pf

V i j

V k


2.24 Rossby Number

To determine the time development we must consider

the acceleration term A in the momentum equations:

1

1

The ratio of the acceleration to Coriolis force is calle

g

g

Du pfv f v v

Dt g x

Dv pfu f u u

Dt g y

2

0 0 0

d

the Rossby number:

DU Dt U L URo

f U f U f L


2.4.3 Hydrostatic Approximation

2 2

The vertical momemtum equation (2.21)

12 cos

Performing scale analysis (Table 2.2), leads to the approximation

1.But there are small variation of p due to advection.

rz

Dw u v pu g F

Dt r z

pg

z

0 0

0

0

0

0

One defines a horizontally averaged pressure p and density , so that

1Apply pertubation theory:

, , , ' , , ,

, , , ' , , , . This leads to

1 '2.29

'

pg

z

p x y z t p z p x y z t

x y z t z x y z t

pg

z


2.5.1 The Continuity EquationEulerian Derivation

Conservation o fmass is a fundamental principle. For a

Eulerian volume element this means that the net inflow

of mass equals the rate of mass increase in the volume.

x

y

z

x y

z -

2

u xu

x

u

x0

y0

z0

2

u xu

x


Eulerian Derivation

Net inflow per unit area

=2 2

Net inflow into the volume through surface y z

= y z

Total net inflow into the volume from all 3 directions

u ux xu u

x x

ux

x

ux

x

u v w

x y

y z y zx x

z

U


Eulerian Derivation

The rate of change of mass within the volume element is

y z. Therefore , or

0 2.30

Since and

,

0 2.31

xt t

t

D

Dt t

D

Dt

Continuity equa

U

U

U U U

U

tion

U


Lagrangian Derivation

Do we get the same result from a Lagragian volume

element? Consider a fixed mass element M moving

with the flow. Since M is conserved

M0

D Dx y z

Dt DtD

x y zDtD x D y D z

y z x z x yDt Dt Dt



,A B

B A

D x xDx Dx D xu u

Dt Dt Dt DtD x

u u uDt

xA xB

UA UB

x



Substitute into

0

0 /

0. For 0 :

0, i.e., the same continuity equation!

D D x D yx y z y z x z

Dt Dt DtD z

x yDt

Dx y z u y z v x z

Dtw x y x y z

D u v w

Dt x y z

D

Dt

U


2.5.3 Scale Analysis of the Continuity Equation

0

00 0 0

00 0 0

0

0 0

0

Set ' , , , :

' ' ' 0

'' 0 for '

1 '' 0

D

Dt tz x y z t

t

t z

w

t z

U U U

U U

U k U

U U


Scale Analysis

0

0 0

00 0

1 '' 0

Scale analysis shows for tropospheric synoptic type

motions B and C. Therefore

0, or 0 2.34

For an fluidincompressibl : 0

0

e

w

t z

A

wz

D

DtD

Dt t

U U

U U

U

A B C

0

0 0.

This is normally not the case for the atmosphere,

except for purely horizontal flow where is a constant.

U U


2.6 Energy Conservation

'The change of internal energy of the system is equal to

the net he

In a fluid at rest, the fi

at added to the system plu

rst l

s the

aw of ther

work done

modyna

by

th

mics state

e sys

s

tem".

For a Lagrangian control volume (moving!) we must

consider the change in mechanical energy to assure the

conservation of energy. Remember the 1st Law of T.D.

is derived from the conservation of energy.


Rate of Change of Energy

2

2

If internal energy of the mass . Then

1 is the total energy of the control volume.

2

1 rate of work by

2

(press. grad gravity+friction+thermal+Coriolis)

We negl

e M V

V e U

DV e U

Dt

ect friction.


Rate of Work Done

The rate at which work is done by the pressure grad. is

(see Fig. 2.7 etc).p V U

The rate at which work is done by the Coriolis force

2

:

0Co F U Ω U U

The rate at which work is done by gravity is V g U

The rate of thermal energy change (radiation, con-

duction, latent heat release) is .V J


1

2

2.35

1 But

2

1 1

2 2

1 since 0

2

DV e

Dt

p V V V J

DV e

Dt

D De V V e

Dt Dt

D D DV e V MDt Dt Dt

U U

U g U

U U

U U U U

U U


Thermal Energy Equation

1

2

12.38

2

DV e p V V V JDt

De p gw V J

Dt

U U U g U

U U U

From 2.8

12

1 12

2

. Into 2.38 :

2.Thermal energy equation 39

Dp

Dt

Dp

Dt

p p gw

Dep J

Dt

Ug Ω U U

U UU g U Ω U U

U g U U

U


Thermodynamic Energy Equation

22 2

We can rewrite 2.39 by using the continuity equation:

10

11

,

For dry air the internal energy is ,

thermodynam

so

ic ener

v

v

D D

Dt Dt

DDe p D D D D

JDt Dt Dt Dt Dt Dt

e c T

DT Dc p JDt Dt

U U

gy equation


Thermodynamics of the Dry Atmosphere

thermodynamic energy equation

RT equation of state. Total derivative:

v

DT Dc p JDt Dtp

Dp D DT D DT Dpp R p R

Dt Dt Dt Dt Dt Dt

Substitute into energy equation above:

2.42

v v

p

DT DT Dp DT Dpc R J c R JDt Dt Dt Dt Dt

DT Dpc J

Dt Dt


Potential Temperature

divide by and use p=RT :

ln ln entropy equation 2.43

p

p

DT Dpc J T

Dt DtD T D p J

c RDt Dt T

For an adiabatic process, J=0.

ln ln 0

ln ln ln ln

Potential temperature

sp p

p

p

pc R c R

T p

R c

s

c D T RD p

D T D p T p

pT

p


Adiabatic Lapse Rate

From ln ln ln ln

1 1 1 1 1 1 1

For an atmosphere for which is constant with height:

2.48

pR c

ss

p

p p p

dp

p RT T p p

p c

T R p T R T gg

z T z p c z T z p c T z T c

T g

z c


2.7.3 Static Stability

If varies with height:

2.49

dp

d

T T g T

z z c z

T

z

If < increases with height, i.e., statically stable.d


Buoyancy Oscillations

2

2

2

2

00

20 0

2

Vertical acceleration is

1 1- -

where p and are the pressure and density of the parcel.

For hydrostatic equilibrium

D zDDw DDt z

Dt Dt Dt

D p pz g g

Dt z z

pg

z

Dz g g g

Dt


Buoyancy …

2

02

We had 2.44 :

......

pR c

spTp

Dz g

Dt

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Reinisch_ASD_85.5151 Chapter 2. Basic Conservation Laws x0x0 y0y0 z0z0 z y x U(x,y,z,t) mass element