1

NOTES FOR THE THEORY OF WKD 263

1. COORDINATE REFERENCE FRAMES

Let r be a position vector of point P that originates from the center of the Earth and let r

be the magnitude of this position vector. Note that zar += where a is the radius of the

Earth and z is the altitude of point P above the surface of the Earth. Let φ be the latitude

(ranging from 2

π− in the Southern Hemisphere to

2

π+ in the Northern Hemisphere) and λ

be the longitude (ranging from 0 at Greenwich to π2 - back at its origin) of point P. R is

defined as a vector perpendicular to the axis of the Earth towards point P at latitude φ.

Figure 1

It is our first intention to express the position and movement of point P as viewed from an

INERTIAL reference frame (movement in the atmosphere as viewed from a fixed point in

space) with unit vectors 321 e,e,e and coordinates x1, x2 and x3 since Newton’s Second

ππππ/2 r

2ππππ

0

R φφφφ

λλλλ

P

2

Law only applies in such an INIRTIAL reference frame. However, as people we are not

viewing Earth from space, but rather rotate with the Earth. It is therefore, secondly, also

convenient to express the position and movement of point P as viewed from a

ROTATING reference frame (movement in the atmosphere as viewed from a fixed point

on the Earth) with unit vectors k,j,i and coordinates x, y and z (Figure 2).

These two reference frames relatively to each other are illustrated in Figure 2:

Figure 2

The observation of longitude might change between the two reference frames as a result

of the rotation of Earth (west to east), while the latitude will not be different as viewed

from the two reference frames because it is not influenced by the rotation of the Earth.

According to Figure 1 the position vector r of point P, relative to coordinate components

in the INERTIAL reference frame, may be written as:

P

x1

x2

x3

r

R

3e

1e

2e

i

j k

λ

φ

y

x

z

3

32a1a esinresincosrecoscosrr φ+λφ+λφ= (1.1)

where aλ refers to the longitude of point P as viewed from the INERTIAL reference

frame. Let λ denote the longitude of a point P in the ROTATING reference frame and let

Ω be the angular speed (speed of rotation = constant ≈ 7.27 x 10-5 rad s-1 = one rotation of

2π every 24 hours) of Earth. The longitude λa changes with time as the Earth rotates

while λ might change as a result of the flow of point P in the ROTATING reference

frame. It therefore follows that )t(t)t(a λ+Ω=λ where Ω is expressed in “radians per

second” meaning that tΩ is the latitudinal angle change as a result of Earth’s rotation.

Equation (1.1) may therefore also be written as

321 esinre)tsin(cosre)tcos(cosrr φ+λ+Ωφ+λ+Ωφ= (1.2)

2. DEFINING THE ANGULAR VELOCITY VECTOR AND VECTOR R

In the INERTIAL reference frame the angular velocity vector of the Earth may be

expressed as follows:

321 ee0e0 Ω++=Ω (2.1)

Note that the vector Ω at any point P is always parallel to the unit vector 3e as defined in

the INERTIAL reference frame. The vector Ω might also be written in terms of the unit

vectors k,j,i of the ROTATING reference frame (see Figure 3). It therefor follows that

k)sin(j)cos( φΩ+φΩ=Ω (2.2)

4

Figure 3

In a similar way the vector R , which is a vector with its origin at the axis of the Earth at

latitude φ and points directly to position P (at a 90º angle with Earth’s axis), can also be written

in terms of the unit vectors k,j,i in the ROTATING reference frame.

Figure 4

From figure 4 it follows that

k)coscosr(j)sincosr(R φφ+φφ−= (2.3)

Ω

j k

3e φ

φ

φ

r

P

j−

k

3e φ

φ

φ R

r

P

5

3. SMALL INCREMENT DISPLACEMENTS

Let δx represent a small longitudinal increment displacement in the i direction (from west to

east) associated with δλ ; δy a small increment latitudinal displacement in the j direction (from

south to north) associated with δφ and δz a small increment displacement associated with δr

(from the surface upwards) in the ROTATING reference frame. Let these increment

displacements take place over an infinitesimal time interval of δt. According to the rule of the

circumference of a circle (circumference = 2πr = angle of path completed multiplied by the radius

of rotation) and R = r cos φ it follows from Figures 1 and 2 that:

φδλ=δ cosrx (3.1)

δφ=δ ry (3.2)

rz δ=δ (3.3)

The speed components u, v, w in the k,j,i directions of the ROTATING reference frame may

therefore be written as:

udt

dcosr

tlimcosr

tcosr

t

x

0t=

λφ=

δ

δλφ=

δ

δλφ=

δ

δ

→δ (3.4)

vdt

dr

tlimr

tr

t

y

0t=

φ=

δ

δφ=

δ

δφ=

δ

δ→δ

(3.5)

wdt

dr

t

rlim

t

r

t

z

0t==

δ

δ=

δ

δ=

δ

δ→δ

(3.6)

Here u, v, w are the speed components in the k,j,i directions as viewed from within the

ROTATING reference frame. By defining U as the wind velocity in the ROTATING

reference frame it follows that

kwjviukdt

drj

dt

dri

dt

dcosrU ++=+

φ+

λφ= (3.7)

6

4. FIRST DERIVATIVE TO OBTAIN VELOCITY

Equation (1.2) gives the components of the position vector r in the INERTIAL reference

frame, which is a function of time as the Earth rotates, and it therefor follows directly that:

321 e)t(sin)t(re))t(tsin()t(cos)t(re))t(tcos()t(cos)t(r)t(r φ+λ+Ωφ+λ+Ωφ=

Let us now define any time derivative operator taken in the INERTIAL reference frame

space as Dt

Da

To obtain the velocity vector we now differentiate the position vector r with respect to

time in the INERTIAL reference frame but with the objective to express r as a position vector in

the ROTATING reference frame

Dt

rDa [ ] 1e)dt

d()tsin(cosr)tcos(

dt

d)sin(r)tcos(cos

dt

dr

λ

+Ωλ+Ω−φ+λ+Ωφ

φ−+λ+Ωφ=

[ ] 2e)dt

d()tcos(cosr)tsin(

dt

d)sin(r)tsin(cos

dt

dr

λ

+Ωλ+Ωφ+λ+Ωφ

φ−+λ+Ωφ+

3edt

dcosrsin

dt

dr

φ

φ+φ+

[ ] 21 e)tcos(e)tsin()(cosrdt

dcosr λ+Ω+λ+Ω−

Ωφ+

λφ=

321 ecose)tsin(sine)tcos(sindt

dr φ+λ+Ωφ−λ+Ωφ−

φ+

321 esine)tsin(cose)tcos(cosdt

drφ+λ+Ωφ+λ+Ωφ+ (4.1)

Replace equations (3.4), (3.5) and (3.6), namely u,v,w which are the wind components in

the ROTATING reference frame, as well as the following unit vectors which are unit

7

vectors in the ROTATING reference frame (Figure 2) relative to the INERTIAL

reference frame.

21 e)tcos(e)tsin(i λ+Ω+λ+Ω−= (4.2)

321 ecose)tsin(sine)tcos(sinj φ+λ+Ωφ−λ+Ωφ−= (4.3)

321 esine)tsin(cose)tcos(cosk φ+λ+Ωφ+λ+Ωφ= (4.4)

Additional notes:

It can be shown that the magnitude of the vectors in equations (4.2), (4.3) and (4.4) is

equal to one meaning that they are unit vectors:

Magnitude of i

11)t(cos)t(sini 22 ==λ+Ω+λ+Ω−=

It can also be shown that these vectors are independent:

For example the independence of vectors i and j

0

sin)tcos()tsin(sin)tcos()tsin(

0))tsin(sin)(tcos())tcos(sin))(tsin((

)ecose)tsin(sine)tcos(sin()e0e)tcos(e)tsin((ji 321321

=

φλ+Ωλ+Ω−φλ+Ωλ+Ω=

+λ+Ωφ−λ+Ω+λ+Ωφ−λ+Ω−=

φ+λ+Ωφ−λ+Ωφ−⋅+λ+Ω+λ+Ω−=⋅

Equation (4.1) then becomes

Dt

rDa [ ] kwjvuiicosr +++φΩ= (4.5)

Equations (3.4), (3.5) and (3.6) indicated that u, v, w are the wind components in the

k,j,i directions as viewed from within the ROTATING reference frame. Since Dt

rDa is

8

the derivative in the INERTIAL reference frame space, the term icosr φΩ in equation

(4.5) must be the angular velocity contribution in the i (west to east) direction as viewed

from the INERSIAL reference frame. This term may be written in a slightly different

way:

r×Ω

φλ+Ωφλ+Ωφ

Ω=

sinr)tsin(cosr)tcos(cosr

00

eee 321

from equations (2.1), (1.2)

[ ] [ ] [ ] 321 e0e)tcos(cosre)tsin(cosr +λ+ΩφΩ+λ+ΩφΩ−=

[ ] [ ] 21 e)tcos(e)tsin(cosr λ+Ω+λ+Ω−φΩ= replace equation (4.2)

icosr φΩ= (4.6)

Replacing the latter into equation (4.5) yields

rkwjvuiDt

rDa ×Ω+++=

and finally (from equation (3.7))

rUUDt

rDa

a ×Ω+== (4.7)

In equation (4.7) we have indicated that the change in the position vector r (velocity

vector aU ), as obtained through differentiating the position in the INIRTIAL reference

frame space, can be written as the velocity in the ROTATING reference frame ( U ) plus

a velocity component in the i direction due to the rotation of the Earth ( r×Ω ).

9

5. SECOND DERIVATIVE TO OBTAIN ACCELERATION

Since we require acceleration for applying the second law of Newton (F = ma per unit

mass) we need to differentiate the velocity vector with respect to time. As before with the

position vector, we will differentiate U which is the velocity vector in the ROTATING

reference frame. From equation (3.7) it follows that

kwjviuU ++=

and therefore

dt

kdwk

dt

dw

dt

jdvj

dt

dv

dt

idui

dt

du

Dt

UDa +++++= (5.1)

Dt

UDa, as defined in equation (5.1), represent differentiation in the INIRTIAL reference

frame space as applied on the velocity vector in the ROTATING reference frame. Take

note that dt

kd,

dt

jd,

dt

id would have been zero in the ROTATING reference frame. Taking

the time derivative of equations (4.2), (4.3) and (4.4), as required in equation 5.1, yields:

kcosr

ujsintan

r

u

dt

id

φΩ+−

φΩ+φ= (5.2)

kr

visintan

r

u

dt

jd

−

φΩ+φ−= (5.3)

jr

vicos

r

u

dt

kd

+

φΩ+= (5.4)

EXERCISE: Derive equations (5.2), (5.3) and (5.4) from equations (4.2), (4.3) and (4.4)

These equations might be written in a slightly different way.

10

Using equation (2.2) yields

kcosjsin

001

sincos0

kji

i φΩ−φΩ=φΩφΩ=×Ω (5.5)

isin

010

sincos0

kji

j φΩ−=φΩφΩ=×Ω (5.6)

icos

100

sincos0

kji

k φΩ=φΩφΩ=×Ω (5.7)

Replace equations (5.5), (5.6) and (5.7) into equations (5.2), (5.3) and (5.4) to obtain

kr

ujtan

r

ui

dt

id

−

φ+×Ω= (5.8)

kr

vitan

r

uj

dt

jd

−

φ−×Ω= (5.9)

jr

vi

r

uk

dt

kd

+

+×Ω= (5.10)

If equations (5.8), (5.9) and (5.10) are replaced in equation (5.1), equation (5.1) becomes

Dt

UD a

kdt

dwj

dt

dvi

dt

du++=

+

+×Ω+

−

φ−×Ω+

−

φ+×Ω+ j

r

vi

r

ukwk

r

vitan

r

ujvk

r

ujtan

r

uiu

11

kdt

dwj

dt

dvi

dt

du++=

+

+

−

φ−+

−

φ+×Ω+×Ω+×Ω+ j

r

vi

r

uwk

r

vitan

r

uvk

r

ujtan

r

uu)k(w)j(v)i(u

but U)kwjviu()kw()jv()iu()k(w)j(v)i(u ×Ω=++×Ω=×Ω+×Ω+×Ω=×Ω+×Ω+×Ω

and therefore

Dt

UDaUk

r

)vu(

dt

dwj

r

wv

r

tanu

dt

dvi

r

wu

r

tanvu

dt

du 222

×Ω+

+

−+

+φ

++

+φ

−=

(5.11)

Equation (5.11) can also be written as the sum of an acceleration term in the ROTATING

reference frame plus the acceleration contribution as a result of Earth’s rotation:

Udt

Ud

Dt

UDa ×Ω+= (5.12)

where

kr

)vu(

dt

dwj

r

wv

r

tanu

dt

dvi

r

wu

r

tanvu

dt

du

dt

Ud 222

+

−+

+φ

++

+φ

−=

12

6. ACCELERATION IN THE INIRTIAL REFERENCE FRAME SPACE

From equations 4.7 and 5.12 it follows that

)rU(Dt

D

Dt

UD aaa ×Ω+=

)r(U2dt

Ud

rUUdt

Ud

UDt

UD

Dt

rD

Dt

UD

aa

aa

×Ω×Ω+×Ω+=

×Ω+×Ω+

×Ω+=

×Ω+=

×Ω+=

(6.1)

The left hand side of equation (6.1) represents the acceleration as viewed from the

INERTIAL reference frame (where the second law of Newton applies), while the right

hand side are written in terms of vectors in the ROTATING reference frame from where

we observe the weather and climate.

7. THE SECOND LAW OF NEWTON AND THE VECTOR FORM OF THE

MOMENTUM EQUATION OF THE ATMOSPHERE

The second law of Newton (F=ma) is only valid in the INERTIAL reference frame, and

therefore (for unit mass; m=1):

Dt

UDF

aa=∑ (7.1)

13

According to equation (6.1) one may also write acceleration in the INIRTIAL reference

frame in terms of components in the ROTATING reference frame. This is more

convenient since we are experiencing atmospheric flow in the ROTATING reference

frame. From equations 6.1 and 7.1 one obtains the MOMENTUM EQUATION for a unit

mass as expressed in terms of vectors in the ROTATING reference frame:

∑ ×Ω×Ω−×Ω−= )r(U2Fdt

Ud (7.2)

8. COORDINATE COMPONENTS OF THE FORCES

In the ROTATING reference frame each force in equation (7.2) may be written in terms

of k,j,i components.

(a) Acceleration in the rotating coordinate system (left hand side of equation 7.2)

kdt

dwj

dt

dvi

dt

du

dt

Ud++= (8.1)

(b) Pressure gradient force (part of ∑F )

We first derive the pressure gradient force in the i (or x) direction. Consider a

cube with an area B over the left vertical panel and an area A over the right

vertical panel. Let the volume of the cube be zyx δδδ . The pressure that pushes

against panel B is p0 and the pressure that pushes against panel B is p0+δp.

Coriolis force

Centripetal acceleration

Pressure gradient force

Gravitational force

Friction and viscosity forces

14

Pressure = Force/ Area and therefore Force=Pressure x Area

Pressure on panel A: xx

pp)pp( oo δ

∂

∂+=δ+ (pressure gradient over distance δx)

Force on panel A: zyxx

ppF oA δδ

δ

∂

∂+−=

Pressure on panel B: op+

Force on panel B: ( ) zypF oB δδ+=

Total force (Fx) in the i direction is therefore

zyxx

pzypzyx

x

pzypFFF ooBAx δδδ

∂

∂=δδ+δδδ

∂

∂−δδ−=+=

But the volume of the cube is ρ

=ρ

===δδδ1m

density

massVzyx

and therefore

x

p1Fx

∂

∂

ρ−=

In a similar way follows that: y

p1Fy

∂

∂

ρ−= and

z

p1Fz

∂

∂

ρ−=

The pressure gradient force in vector and component format is therefore:

kz

p1j

y

p1i

x

p1p

1

∂

∂

ρ−

∂

∂

ρ−

∂

∂

ρ−=∇

ρ− (8.2)

A B

δx

δy

δz p0 p0+δp

15

(c) Gravitational force (part of ∑F )

The gravitational force may be written as ∗g (8.3)

which is a vector that points towards the centre of the Earth. This vector is not

necessarily parallel to the unit vector k since it is not always perpendicular to the

surface of the Earth. The reason for this is that the Earth is not a perfect circle, but

rather a sphere:

(d) Frictional forces (part of ∑F )

kFjFiFF rzryrxr ++= __________(8.4)

(e) Coriolis force

From equations (2.2) and (3.7) it follows that

k)cos2u(j)sin2u(i)sin2vcos2w(

wvu

sincos0

kji

2U2

φΩ+φΩ−φΩ+φΩ−=

φΩφΩ−=×Ω− (8.5)

∗g

16

(f) Centripetal force (and gravity force)

From equations (2.2) , (4.6) and (2.3) it follows that

R

k)coscosr(j)sincosr(

k)coscosr(j)sincosr(i)0(

00cosr

sincos0

kji

)r(

2

2

22

Ω−=

φφ+φφ−Ω−=

φφΩ−φφΩ+=

φΩ

φΩφΩ=×Ω×Ω−

If we combines the vector with the same magnitude but opposite direction of the

centripetal force (called the centrifugal force) with the gravitational force

(equation 8.3), we get the gravity force ( Rgg 2* Ω+= ). Unlike the gravitational

force vector, the gravity force vector will always be perpendicular to the spherical

surface of the Earth and will therefore always be parallel to the unit vector k ,

although in the opposite direction.

The gravity force which is the sum, or combination, of the gravitational and

centrifugal forces may therefore be written as:

kgg −= (8.6)

where Rg)r(gg 2** Ω+=×Ω×Ω+=

∗g

g

R2Ω

17

9. COORDINATE COMPONENTS OF THE FORCES

From equation 7.2 the MOMENTUM EQUATION for a unit mass is:

∑ ×Ω×Ω−×Ω−= )r(U2Fdt

Ud

which in coordinate components (when replacing equations 8.1, 8.2, 8.4, 8.5 and 8.6)

may be written as:

i - direction : rxFx

p1

a

tanuv

a

uwcosw2sinv2

dt

du+

∂

∂

ρ−=

φ−+φΩ+φΩ− (9.7)

j - direction : ry

2

Fy

p1

a

tanu

a

vwsinu2

dt

dv+

∂

∂

ρ−=

φ−+φΩ+ (9.8)

k - direction : rz

22

Fgz

p1

a

)vu(cosu2

dt

dw+−

∂

∂

ρ−=

+−φΩ− (9.9)

or in vector format:

rFU2gp1

dt

Ud+×Ω−+∇

ρ−= (9.10)

18

Simplification of the momentum equation (scaling for synoptic systems)

On the synoptic scale we consider systems that have a horizontal scale of in the order of

1000km and a vertical scale of in the order of 10km (approximated depth of the

troposphere). If we also consider mid-latitude systems the scales might be summarised as

follows:

L: Horizontal length scale = 10 6 m (1000km)

H: Vertical depth scale = 10 4 m (10km)

U: Horizontal velocity scale = 10 m.s -1

(36 km.h -1

)

W: Vertical velocity scale = 10 -2

m.s -1

(1 cm.s -1

)

δP/ρ: Horizontal pressure fluctuation scale = 10 3 m

2.s

-2

T: Time scale (L/U) = 10 6/ 10 = 10

5s

ν: Kinematic viscosity coefissient (for friction) = 1.46 x 10 -5

m 2.s

-1

2Ω: 2 x Speed of rotation of the Earth = 1.4 x 10 -4

rad.s -1

g: Gravity = 10 m.s -2

a: Radius of the Earth = 6.37 x 10 6 m

In the synoptic scale small terms (grey) might be ignored:

rxFx

p1

a

tanuv

a

uwcosw2sinv2

dt

du+

∂

∂

ρ−=

φ−+φΩ+φΩ−

U2/L 2ΩU 2ΩW UW/a U

2/a δP/Lρ νU/H

2

(10 -4

) (10 -3

) (10 -6

) (10 -8

) (10 -5

) (10 -3

) (10 -12

)

ry

2

Fy

p1

a

tanu

a

vwsinu2

dt

dv+

∂

∂

ρ−=

φ−+φΩ+

U2/L 2ΩU UW/a U

2/a δP/Lρ νU/H

2

(10 -4

) (10 -3

) (10 -8

) (10 -5

) (10 -3

) (10 -12

)

rz

22

Fgz

p1

a

)vu(cosu2

dt

dw+−

∂

∂

ρ−=

+−φΩ−

UW/L 2ΩU U2/a δP/Hρ g νW/H

2

(10 -7

) (10 -3

) (10 -5

) (10) (10) (10 -15

)

19

10 THE CONTINUITY EQUATION

In the previous section we derived the momentum equation that addressed the

CONSERVATION OF MOMENTUM in the atmosphere. The second fundamental

conservation principle is the CONSERVATION OF MASS in the atmosphere. The

mathematical relationship that expresses CONSERVATION OF MASS is called the

continuity equation.

Consider a fixed (stationary) unit volume element zyx δδδ with density ρ and mass mδ .

In order to conserve mass within the volume, the following must apply:

RATE OF MASS INFLOW ACROSS THE BOUNDARIES OF THE VOLUME

= RATE OF MASS INCREASE IN THE VOLUME

1) RATE OF MASS INFLOW ACROSS THE BOUNDARIES OF THE VOLUME

We know that the RATE OF MASS INFLOW = MOMENTUM TRANSFER which, for

a unit volume, may be expressed as:

kwjviu)1(UUzyxUm ρ+ρ+ρ=ρ=δδρδ=δ where kwjviuU ++= .

MOMENTUM TRANSFER in the i direction across area B may therefore be written as:

δx

δy

δz zyx)u(

xu δδ

δρ

∂

∂+ρ A B ( ) zyu δδρ

20

( ) zyu δδρ (10.1)

Momentum transfer in the i direction over panel A, which is distance xδ from panel B,

therefore becomes

zyx)u(x

u δδ

δρ

∂

∂+ρ (10.2)

The net flow into the unit volume owing to the i velocity component then becomes

( ) ( ) zyxux

zyx)u(x

uzyu δδδρ∂

∂−=δδ

δρ

∂

∂+ρ−δδρ (10.3)

A similar expression can be derived for momentum transfer across areas in the j and k

directions, meaning that the net mass flux over all areas becomes

( ) ( ) ( ) zyxwz

vy

ux

δδδ

ρ

∂

∂+ρ

∂

∂+ρ

∂

∂−

and for unit volume

( ) ( ) ( )

( )U

1.wz

vy

ux

ρ⋅∇−=

ρ

∂

∂+ρ

∂

∂+ρ

∂

∂−

(10.4)

2) RATE OF MASS INCREASE IN THE VOLUME

The RATE OF MASS INCREASE IN THE VOLUME may be expressed as:

21

t)zyx(

tt

m

∂

ρ∂=δδρδ

∂

∂=

∂

∂ for the unit volume (10.5)

Setting equation (10.4) equal to equation (10.5) yields - as originally defined

( )t

U∂

ρ∂=ρ⋅∇−

( ) 0Ut

=ρ⋅∇+∂

ρ∂ (10.6)

Equation (10.7) is known as the mass divergence form of the continuity equation for a

stationary Eulerian unit volume. One can also write the continuity equation in terms of

velocity divergence looking at Lagrangian volume that moves in the atmosphere:

( ) ( ) 0UUUdt

dU

t=⋅∇ρ+ρ∇⋅+

ρ∇⋅−

ρ=ρ⋅∇+

∂

ρ∂

and therefore

0Udt

d1=⋅∇+

ρ

ρ (10.7)

Remember, as earlier defined in Vector

Algebra, that t∂

∂ refers to temporal

change in a stationary Eulerian volume

while dt

d refers to changes in a moving

Lagrangian volume

22

11 THE THERMODYNAMIC ENERGY EQUATION

The final conservation equation is known as the the thermodynamic energy equation that

describes the CONSERVATION OF ENERGY in the atmosphere.

The total thermodynamic energy of a moving (Lagrangian) control volume of air (also a

mass element Vm ρδ=δ ) in the atmosphere where mass is conserved may be expressed

as the sum of its internal heating energy and the kinetic energy as a result of macroscopic

(large-scale) flow.

Therefore:

RATE OF CHANGE IN THE TOTAL THERMODYNAMIC ENERGY = RATE OF

INTERNAL HEATING IN THE VOLUME OF AIR + THE RATE AT WHICH WORK

IS DONE ON THE VOLUME OF AIR BY EXTERNAL FORCES.

The external forces may be defided into (1) SURFACE FORCES (pressure and viscosity

on the surface boundary of the volume) and BODY FORCES (gravity and Corioilis

force).

Let

δme = internal energy of a mass element δm with volume Vδ and density ρ

δm½ UU ⋅ = kinetic energy on a mass element δm as a result of external forces

Energy =

Internal energy

External surface forces

External body forces

23

The RATE OF CHANGE IN THE TOTAL THERMODYNAMIC ENERGY of a mass

element can therefore be written as:

ρδ⋅+=

⋅+δ V)UU2

1e(

dt

d)UU

2

1e(m

dt

d (11.1)

1) The rate of internal heating on a mass element due to radiation, conduction and

latent heat release, may be written as:

mJδ = )V(J ρδ (11.2)

where J is the rate of heating per unit mass

2) The rate of work done on a mass element by an external surface PRESSURE

FORCE, may be written as

UF ⋅ = u)zyp( δδ = zy)pu( δδ

The rate of work done by the external surface PRESSURE FORCE in the i

direction on area B may therefore be written as:

( ) zypu B δδ

δx

δy

δz A B zyx)pu(

x)pu( BB δδ

δ

∂

∂+− zy)pu( B δδ

24

The rate of work done by the external surface PRESSURE FORCE in the i

direction on area A, which is distance xδ from panel B, therefore becomes

zyx)pu(x

)pu( B δδ

δ

∂

∂+−

The net rate of work done by the external surface PRESSURE FORCE in the i

direction then becomes

( ) ( ) ( ) Vpux

zyxpux

zyx)pu(x

)pu(zypu BB δ∂

∂−=δδδ

∂

∂−=δδ

δ

∂

∂+−δδ

A similar expression can be derived for the rate of work done by external surface

PRESSURE FORCES in the j and k directions, meaning that the net rate of work

done on all areas becomes

( ) ( ) ( ) zyxpwz

pvy

pux

δδδ

∂

∂+

∂

∂+

∂

∂−

( ) VUp δ⋅∇−= (11.3)

3) The rate of work done on the body of a mass element by the external CORIOLIS

FORCE (equation 7.2) may be written as:

UF ⋅ = )UU2(m ⋅×Ω−δ = )UU2(m ×⋅Ω−δ = 0 (11.4)

(because wind Coriolis force)

25

4) The rate of work done on the body of a mass element by the GRAVITY FORCE

(equation 8.6) may be written as:

UF ⋅ = Ugm ⋅δ = Ug)V( ⋅ρδ = VUg ρδ⋅ = Vgwρδ− (11.5)

where kgg −= is the gravity force par unit mass

From equations (11.2), (11.3), (11.4) and (11.4), the RATE OF CHANGE IN THE

TOTAL THERMODYNAMIC ENERGY (equation 11.1) can now be written as:

( ) VgwVUpVJV)UU2

1e(

dt

dρδ−δ⋅∇−ρδ=

ρδ⋅+ (11.6)

Apply the chain rule of differentiation

( ) VgwVUpVpUVJ)V(dt

d)UU

2

1e()UU

2

1e(

dt

dV ρδ−δ⋅∇−δ∇⋅−ρδ=ρδ⋅++⋅+ρδ (11.7)

As a result of the conservation of mass, equation (10.7), 0)m(dt

d)V(

dt

d=δ=ρδ and divide

with Vδ

ρ−∇⋅−⋅∇−ρ=⋅ρ+ρ gwpUUpJ)UU2

1(

dt

d

dt

de (11.8)

Terms above brackets is the same as in equation (A)

in Appendix A which is the balance of mechanical energy.

The remaining terms represent the thermal energy by internal heating.

If we replace equation (A) in Appendix A in equation (11.8), equation (11.8) becomes

26

UpJdt

de⋅∇−ρ=ρ (11.9)

which is the thermal energy equation that can also be written as

Up

dt

deJ ⋅∇

ρ+= (11.10)

• We finally replace the following terms in equation (11.10)

Let the internal energy per unit mass Tce v= which yields

dt

dTc

dt

)Tc(d

dt

dev

v ==

where 11

v KJkg717c −−= (specific heat at constant volume Vδ )

• According to continuity equation (10.7) dt

d1U

ρ

ρ−=⋅∇ .

Replacing this into Up

⋅∇ρ

yields

dt

d1

dt

d)(

dt

dp

dt

dp

dt

dp 12

2

α=

ρ=ρ=

ρρ−=

ρ

ρ− −−

where ρ

=α1

to obtain the general form of the THERMODYNAMIC ENERGY EQUATION:

dt

dp

dt

dTcJ v

α+= (11.11)

The second term on the right that describes the rate of work done (per unit mass)

represents a conversion between thermal and mechanical energy. This conversion

process enables the solar heat energy to drive motions in the atmosphere.

27

APPENDIX A

As indicated in equation (9.10) the MOMENTUM equation in vector format can be

written as:

rFU2gp1

dt

Ud+×Ω−+∇

ρ−=

Neglect rF because it is regarded as small, and apply ( ⋅U )

UU2UgUp1

dt

UdU ⋅×Ω−⋅+⋅∇

ρ−=⋅

Multiply with ρ and let gwUg −=⋅ ; 0UU2UU2 =×⋅Ω−=⋅×Ω− ; dt

UdU2

dt

)UU(d⋅=

⋅

gwUpdt

)UU(d

2

1ρ−⋅∇−=

⋅ρ

gwpU)UU2

1(

dt

dρ−∇⋅−=⋅ρ (A)

which represents the balance of mechanical energy as a result of the movement of the

volume, which is known as the mechanical energy equation.