AOSS 321, Winter 2009 Earth System Dynamics Lecture 5 1/22

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```							        AOSS 321, Winter 2009
Earth System Dynamics

Lecture 5
1/22/2009

Christiane Jablonowski       Eric Hetland
cjablono@umich.edu       ehetland@umich.edu
734-763-6238             734-615-3177
Class News
• Class web site:
https://ctools.umich.edu/portal
• HW 1 due today
• Homework #2 posted today, due on Thursday
(1/29) in class
• Our current grader is Kevin Reed
(kareed@umich.edu)
• Office Hours
– Prof. Jablonowski, 1541B SRB: Tuesday after
class 12:30-1:30pm, Wednesday 4:30-5:30pm
– Prof. Hetland, 2534 C.C. Little, TBA
Today’s class

•   Definition of the the Total (Material) Derivative
•   Lagrangian and Eulerian viewpoints
•   Fundamental forces in the atmosphere:
Surface forces:
–…
Total variations
Consider some parameter, like temperature, T
Δx

y

Δy
x
If we move a parcel in time Δt

Using Taylor series expansion

T      T      T      T         Higher
T     t     x     y     z     Order
t      x      y      z         Terms

Assume increments over Δt are small, and
ignore Higher Order Terms
Total derivative
Total differential/derivative of the temperature T,
T depends on t, x, y, z

T      T      T      T
T     t     x     y     z
t      x      y      z

Assume increments over Δt are small

Total Derivative
Divide by Δt

T T T x T y T z
            
t t x t y t z t

Take limit for small Δt

dT T T dx T dy T dz
            
dt t x dt y dt z dt
Total Derivative
Introduction of convention of d( )/dt ≡ D( )/Dt

DT T T Dx T Dy T Dz
            
Dt t x Dt y Dt z Dt

This is done for clarity.

By definition:
Dx      Dy      Dz
 u,     v,    w
Dt      Dt      Dt
u,v,w: these are the velocities
Definition of the Total Derivative

DT T    T    T    T
   u    v    w
Dt t    x    y    z

The total derivative is also
called material derivative.

D()
describes a ‘Lagrangian viewpoint’
Dt
()    ()    ()    ()
u     v     w     describes an ‘Eulerian viewpoint’
t     x     y     z
Lagrangian view
Position vector at different times

Consider fluid parcel moving along some trajectory.
Lagrangian Point of View
• This parcel-trajectory point of view, which
follows a parcel, is known as the Lagrangian
point of view.
– Useful for developing theory
– Requires considering a coordinate system
for each parcel.
– Very powerful for visualizing fluid motion
Lagrangian point of view:
Eruption of Mount Pinatubo
• Trajectories trace the motion of individual fluid
parcels over a finite time interval
• Volcanic eruption in 1991 injected particles into
the tropical stratosphere (at 15.13 N, 120.35 E)
• The particles got transported by the atmospheric
flow, we can follow their trajectories
• Mt. Pinatubo, NASA animation
• Colors in animation reflect the atmospheric height of
the particles. Red is high, blue closer to the surface.
• This is a Lagrangian view of transport processes.
Global wind systems
• General Circulation of the Atmosphere
Zonally averaged circulation
• Zonal-mean annual-mean zonal wind u


Pressure (hPa)
Eulerian view
Now we are going to really think about fluids.

Could sit in one place and watch parcels go by.
How would we quantify this?
Eulerian Point of View
• This point of view, where is observer sits at a
point and watches the fluid go by, is known as
the Eulerian point of view.
– Useful for developing theory
– Looks at the fluid as a field.
– Requires considering only one coordinate system
for all parcels
– Easy to represent interactions of parcels through
surface forces
– A value for each point in the field – no gaps or
bundles of “information.”
An Eulerian Map
Consider some parameter, like
temperature, T

y

x

DT
 Material derivative, T change following the parcel
Dt
Consider some parameter, like
temperature, T

y

x

T
 Local T change at a fixed point
t
Consider some parameter, like
temperature, T

y

x

T    T    T
v  T  u    v    w
x    y    z


Consider some parameter, like
temperature, T

T
u                   y
x

T                        x
v
y

v  T  0 : warm air advection
v  T  0 : cold air advection
Advection of cold or warm air
• Temperature advection: v  T
• Imagine the isotherms are oriented in the E-W
direction         warm


u

y
X
cold
• Draw the horizontal temperature gradient vector!
• pure west wind u > 0, v=0, w=0: Is there
temperature advection? If yes, is it cold or warm
Advection of cold or warm air
• Temperature advection: v  T
• Imagine the isotherms are oriented in the E-W
direction
cold

v   

y
warm
X
• Draw the gradient of the temperature (vector)!
• pure south wind v > 0, u=0, w=0: Is there
temperature advection? If yes, is it cold or warm
Advection of cold or warm air
• Temperature advection: v  T
• Imagine the isotherms are oriented as

cold

u
warm
y
• Draw the horizontal temperature gradient!       X
• pure west wind u > 0, v=0, w=0: Is there
temperature advection? If yes, is it cold or warm
Summary:
Local Changes & Material Derivative

T DT       T      T    T
     u    v      w
t    Dt    x      y    z
T       DT
         v  T
t       Dt
at a fixed location
                    Total change along
a trajectory
Summary: For 2D horizontal flows
T DT       T      T
    u      v
t   Dt     x      y
T DT
     vh   hT
t   Dt
u
with v h    horizontal wind vector and
v 
                
x 
 h     horizontal gradient operator
               
y 
DT
if     0  Conservation of T
Dt
T
if     0  Steady state is reached
t

Remember: we talked about the conservation of
money

Conservation principle is important for
tracers in the atmosphere that do not have
sources and sinks
Class exercise
• The surface pressure decreases by 3 hPa per
180 km in the eastward direction.
• A ship steaming eastward at 10 km/h measures
a pressure fall of 1 hPa per 3 hours.
• What is the pressure change on an island that
the ship is passing?
N
NW        NE

Directions:   W               E

SW        SE
S
Food for thought
• Imagine a different situation.
• The surface pressure decreases by 3 hPa per
180 km in the north-east direction.
• Thus:

Low p

u

High p
What are the fundamental forces in
the Earth’s system?
•   Gravitational force
•   Viscous force
•   Apparent forces: Centrifugal and Coriolis
•   Can you think of other classical forces and
would they be important in the Earth’s system?

• Total Force is the sum of all of these forces.
A particle of atmosphere
r ≡ density = mass
per unit volume (V)

z    V = xyz

m = rxyz
---------------------------------
y          p ≡ pressure =
force per unit area
x
acting on the particle of
atmosphere
Check out Unit 6, frames 7-13:
http://www.atmos.washington.edu/2005Q1/101/CD/MAIN3.swf
(x0, y0, z0)            p0 = pressure at (x0, y0, z0)

z
Pressure at the ‘wall’:
p x
p .0
p  p0 
x 2


higher order terms
y
Remember the Taylor
x                      series expansion!

x axis
Pressure at the ‘walls’
p x                                        p x
p  p0                                     p  p0        
x 2                                         x 2
higher order terms                           higher order terms

z
p . 0   

y
F
x                       remember: p 
A
x axis
(ignore higher order terms)
     p x 
FBx  p0         
 yz
     x 2                     Area of side A:
A     yz

.
z
B              y
Area of side B:
yz                  x                     p x 
FAx  p0         
 yz
     x 2 
x axis                              Watch out for the + and -
directions!
Pressure gradient force (4):Total x force
Fx  FBx  FAx
     p x             p x 
 p0         
 yz  p0         
 yz
     x 2              x 2 
p
  xyz
x
We want force per unit mass
Fx  p        
               xyz rxyz
 
m  x         
1 p

r x

1 p p p
F / m   ( i  j  k)
r x y z
z k
1
F / m   p           y j   x i
r
Class exercise
Compute the pressure gradient force at sea level in x
and y direction at 60°N          Assume constant
Isobars with contour              density r = 1.2 kg/m3
interval p = 5 hPa                  and radius
a = 6371 km

L
1000 hPa                          = 20º = /9
x = a cos 

Low pressure system                  = 20º = /9
at 60°N                             y = a 
Class exercise
Compute
990
the pressure            1016
at the surface                 1008
1000

Contour
1041
interval:
1008
4 hPa                                 1034

Density?
1012
NCAR
forecasts
Our momentum equation so far

dv    1
  p  other forces
dt    r

Here, we use the text’s convention that the velocity is

v  u,v,w
Highs and Lows

Pressure gradient force tries to eliminate
the pressure differences

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