chapter 2. basic conservation laws
DESCRIPTION
z. U (x,y,z,t). mass element. z 0. y 0. y. x 0. x. Chapter 2. Basic Conservation Laws. 2.1 Total Differentiation ( in the Lagrangian frame). Local Temperature Change and Temperature Advection. Example for Advection. p. p(x). 300 Pa. u=10km/h. x. East. West. 180 km. - PowerPoint PPT PresentationTRANSCRIPT
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Lagrangian Derivation
,A B
B A
D x xDx Dx D xu u
Dt Dt Dt DtD x
u u uDt
xA xB
UA UB
x
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Lagrangian Derivation
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
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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
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Reinisch_ASD_85.515 26
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
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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.
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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.
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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
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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
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Reinisch_ASD_85.515 31
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
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Reinisch_ASD_85.515 32
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
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Reinisch_ASD_85.515 33
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
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Reinisch_ASD_85.515 34
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
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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
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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
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Reinisch_ASD_85.515 37
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
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Reinisch_ASD_85.515 38
Buoyancy …
2
02
We had 2.44 :
......
pR c
spTp
Dz g
Dt