Transcript AOSS_401_20070907_L02_Forces

AOSS 401, Fall 2007
Lecture 2
September 7, 2007
Richard B. Rood (Room 2525, SRB)
[email protected]
734-647-3530
Derek Posselt (Room 2517D, SRB)
[email protected]
734-936-0502
Class News
• Ctools site (AOSS 401 001 F07)
– Calendar (completed for whole semester)
– Syllabus
– Lectures
• Posted on day of
– Homework (and solutions)
• Homework has been posted
– Under “resources” in homework folder
• Due next Wednesday (September 12, 2007)
Class news: Schedule issues
• Currently 4.5 hours are scheduled for a
4.0 hour course. (So we have some
flexibility; we can “cancel” 4 classes)
– There will be no class on September 14
– There will be no class on October 12
– There will be no class on November 21
– When to schedule final exam?
Weather
• National Weather Service
– http://www.nws.noaa.gov/
– Model forecasts:
http://www.hpc.ncep.noaa.gov/basicwx/day07loop.html
• Weather Underground
– http://www.wunderground.com/cgibin/findweather/getForecast?query=ann+arbor
– Model forecasts:
http://www.wunderground.com/modelmaps/maps.asp
?model=NAM&domain=US
Outline
• Pressure gradient force
• Gravitational Force
• Viscous force
• Centrifugal Force
• Coriolis Force
Should be review. So we are going fast.
You have the power to slow us down.
Some basics of the atmosphere
atmosphere: depth ~ 1.0 x 105 m
Mountain: height ~ 5.0 x 103 m
Ocean
Land
Earth: radius ≡ a = 6.37 x 106 m
Biosphere
Some basics of the atmosphere
Troposphere
------------------ ~ 2
Mountain
Troposphere
------------------ ~ 1.6 x 10-3
Earth radius
Troposphere: depth ~ 1.0 x 104 m
This scale analysis tells us that the troposphere is thin relative to the size of
the Earth and that mountains extend half way through the troposphere.
Newton’s Law of Motion
F = ma
Force = mass x acceleration
In general we will work with force per unit mass; hence,
a = F/m
And with the definition of acceleration
Bold will represent vectors.
Newton’s Law of Motion
dv
F/m
dt
Which is the vector form of the momentum equation.
(Conservation of momentum)
What are the forces?
•
•
•
•
•
Pressure gradient force
Gravitational force
Viscous force
Apparent forces
Can you think of other classical forces and
would they be important in the Earth’s
atmosphere?
• Total Force is the sum of all of these forces.
Newton’s Law of Motion
dv 1
  Fi
dt m i
Where i represents the different types of forces.
How do we express the forces?
• In general, we assume the existence of an
idealized parcel or “particle” of fluid.
• We calculate the forces on this idealized parcel.
• We take the limit of this parcel being
infinitesimally small.
– This yields a continuous, as opposed to discrete,
expression of the force.
• Use the concept of the continuum to extend this
notion to the entire fluid domain.
An intrinsic assumption
• There is an equation of state that
describes the thermodynamic properties of
the fluid, the air.
A particle of atmosphere
r ≡ density =
mass per unit volume (DV)
Dz
DV = DxDyDz
m = rDxDyDz
-------------------------------------
Dy
z k
Dx
y j
x i
p ≡ pressure =
force per unit area
acting on the particle of
atmosphere
Pressure gradient force (1)
(x0, y0, z0)
p0 = pressure at (x0, y0, z0)
Dz
p = p0 + (∂p/∂x)Dx/2 +
higher order terms
.
Dy
Dx
x axis
Pressure gradient force (2)
p = p0 - (∂p/∂x)Dx/2 +
higher order terms
Dz
.
p = p0 + (∂p/∂x)Dx/2 +
higher order terms
Dy
Dx
x axis
Pressure gradient force (3)
(ignore higher order terms)
FBx = (p0 - (∂p/∂x)Dx/2) (DyDz)
Dz
.
FAx = - (p0 + (∂p/∂x)Dx/2) (DyDz)
A
B
Dy
Dx
x axis
Pressure gradient force (4)
Total x force
Fx = FBx + FAx
= (p0 - (∂p/∂x)Dx/2) (DyDz) - (p0 + (∂p/∂x)Dx/2) (DyDz)
= - (∂p/∂x)(DxDyDz)
We want force per unit mass
Fx/m = - 1/r (∂p/∂x)
Vector pressure gradient force
1 p p
p
F/m   ( i
j  k)
r x y
z
z k
F/m  
1
r
p
y j
x i
Viscous force (1)
Derivation is at end of lecture or in the text.
• There is in a fluid friction, which resists the
flow. It is dissipative, and if the fluid is not
otherwise forced, It will slow the fluid and
bring it to rest. Away from boundaries in
the atmosphere this frictional force is often
small, and it is often ignored. (We will
revisit this as we learn more.)
• Close to the boundaries, however, we
have to consider friction.
Viscous force (2)
velocity ≡ u
m/sec
Ocean
Land
Biosphere
velocity must be 0 at surface
Viscous force (3)
velocity ≡ u
m/sec
Ocean
Land
Biosphere
Velocity is zero at the surface; hence,
there is some velocity profile.
Viscous force (4)
(How do we think about this?)
The drag on the moving plate is the same as the force required to
keep the plate moving. It is proportional to the area (A), proportional
to the velocity of the plate, and inversely proportional to the
distance between the plates; hence,
u(h) = u0
h
u(z)
F = μAu0/h
u(0) = 0
Proportional usually means we assume linear relationship. This is a
model based on observation, and it is an approximation. This is often
said to be “Newtonian.” The constant of proportionality assumes some
physical units. What are they?
Viscous force (5)
 ≡ m/r  kinematic viscosity coefficient
F
2
  (u)
m
Where Laplacian is operating on velocity
vector ≡ u = (u, v, w)
Surface forces
• Pressure gradient force and the viscous force
are examples of a surface force.
• Surface forces are proportional to the area of the
surface of our particle of atmosphere.
• Surface forces are independent of the mass of
the particle of atmosphere.
• They depend on characteristics of the particle of
atmosphere; characteristics of the flow.
Highs and Lows
Motion initiated by pressure gradient
Opposed by viscosity
Where’s the low pressure?
Geostrophic and observed wind 1000 mb (ocean)
Body forces
• Body forces act on the center of mass of
the parcel of fluid.
• Magnitude of the force is proportional to
the mass of the parcel.
• The body force of interest to dynamic
meteorology is gravity.
Newton’s Law of Gravitation
Gm1m2 r
F
2
r
r
Newton’s Law of Gravitation: The force between any
two particles having masses m1 and m2 separated
by distance r is an attraction acting along the line
joining the particles and has the magnitude
proportional to G, the universal gravitation constant.
Gravitational Force
Gravitational force for dynamic meteorology
GMm r
F 2
r r
Newton’s Law of Gravitation:
M = mass of Earth
m = mass of air parcel
r = distance from center (of mass) of Earth
directed down, towards Earth, hence - sign
Adaptation to dynamical meteorology
GM
g0  2
a
r az
a is radius of the Earth
z is height above the Earth’s surface
Can we ignore z, the height above the surface? How would you make that
argument?
Gravity for Earth
a
mg0
a2
=g0
a
Gravitational force per unit mass
2
F
a
r
  g0
2
m
(a  z ) r
Our momentum equation
2
du
1
a
r
2
  p  (u)  g0
dt
r
(a  z ) 2 r
+ other forces
Now using the text’s convention that the velocity is
u = (u, v, w).
Apparent forces
Back to Basics:
Newton’s Laws of Motion
• Law 1: Bodies in motion remain in motion with
the same velocity, and bodies at rest remain at
rest, unless acted upon by unbalanced forces.
• Law 2: The rate of change of momentum of a
body with time is equal to the vector sum of all
forces acting upon the body and is the same
direction.
• Law 3: For every action (force) there is and
equal and opposite reaction.
Back to basics:
A couple of definitions
• Newton’s laws assume we have an
“inertial” coordinate system; that is, and
absolute frame of reference – fixed,
absolutely, in space.
• Velocity is the change in position of a
particle (or parcel). It is a vector and can
vary either by a change in magnitude
(speed) or direction.
Apparent forces:
A mathematical approach
• Non-inertial, non-absolute coordinate
system
Two coordinate systems
Can describe the velocity
and forces (acceleration)
in either coordinate
system.
z
y
x
dx dx '
or
dt
dt
One coordinate system related to another by:
x'  Ax (t ) x  Bx (t ) y  C x (t ) z
y '  Ay (t ) x  B y (t ) y  C y (t ) z
z '  Az (t ) x  Bz (t ) y  C z (t ) z
Velocity (x’ direction)
x'  Ax (t ) x  Bx (t ) y  C x (t ) z
d ( Bx (t ))
d (C x (t ))
dx'
dx
dy
dz d ( Ax (t ))
 Ax (t )  Bx (t )  C x (t ) 
x
y
z
dt
dt
dt
dt
dt
dt
dt
So we have the velocity relative to the coordinate system
and the velocity of one coordinate system relative to the
other.
This velocity of one coordinate system relative to the other
leads to apparent forces. They are real, observable forces
to the observer in the moving coordinate system.
Two coordinate systems
z’ axis is the same as z,
and there is rotation of
the x’ and y’ axis
z’
z
y
y’
x
x’
One coordinate system related to another by:
x'  x cos(t )  y sin( t )
y '   x sin( t )  y cos(t )
z'  z
2

T
T is time needed to complete rotation.
Acceleration (force) in rotating coordinate system
d 2 x'
dx
dy
2


2

(
sin(

t
)

cos(

t
))


( x cos(t )  y sin( t ))
2
dt
dt
dt
d 2 y'
dx
dy
2


2

(
cos(

t
)

sin(

t
))


( x sin( t )  y cos(t ))
2
dt
dt
dt
d 2 z'
0
2
dt
The apparent forces that are proportional to rotation and the
velocities in the inertial system (x,y,z) are called the Coriolis
forces.
The apparent forces that are proportional to the square of the
rotation and position are called centrifugal forces.
Apparent forces:
A physical approach
Circle Basics
ω
r (radius)
θ
Arc length ≡ s = rθ
ds
d
r
dt
dt
d

dt
ds
 v 
dt
... Magnitude
s = rθ
Centrifugal force:
Treatment from Holton
dv
d r
  v
dt
dt r
ω
r (radius)
Δv
Δθ
...
Magnitude
Centrifugal force: for our purposes
dv
d r
  v
dt
dt r
d

dt
 v  r
dv
2
  r
dt
Now we are going to think about
the Earth
• The preceding was a schematic to think
about the centrifugal acceleration problem.
Note that the r vector above and below are
not the same!
What direction does the Earth’s
centrifugal force point?
Ω
Ω2R
R
Earth
What direction does gravity point?
Ω
F
a2 r
  g0
2
m
(a  z ) r
R
Earth
a
What direction does the Earth’s
centrifugal force point?
Ω
Ω2R
R
Earth
So there is a component
that is in the same
coordinate direction as
gravity (and local vertical).
And there is a component
pointing towards the equator
We are now explicitly
considering a coordinate
system tangent to the
Earth’s surface.
What direction does the Earth’s
centrifugal force point?
Ω
Ω2R
R
Φ = latitude
Earth
So there is a component
that is in the same
coordinate direction as
gravity:
~ aΩ2cos2(f)
And there is a
component pointing
towards the equator
~ - aΩ2cos(f)sin(f)
So we re-define gravity as
F
a2
r
2
2
 ( g 0
 a cos (f ))
2
m
(a  z )
r
F
r
 g
m
r
What direction does the Earth’s
centrifugal force point?
Ω
Ω2R
R
Earth
And there is a
component pointing
towards the equator.
The Earth has bulged to
compensate for the
equatorward
component.
Hence we don’t have to
consider the horizontal
component explicitly.
Centrifugal force of Earth
• Vertical component incorporated into redefinition of gravity.
• Horizontal component does not need to be
considered when we consider a coordinate
system tangent to the Earth’s surface,
because the Earth has bulged to
compensate for this force.
• Hence, centrifugal force does not appear
EXPLICITLY in the equations.
Apparent forces:
A physical approach
• Coriolis Force
• http://climateknowledge.org/figures/AOSS
401_coriolis.mov
Next time
• Coriolis Force
– Read and re-read the section in the text.
• Pressure as a vertical coordinate
– Geopotential
Weather
• National Weather Service
– http://www.nws.noaa.gov/
– Model forecasts:
http://www.hpc.ncep.noaa.gov/basicwx/day07loop.html
• Weather Underground
– http://www.wunderground.com/cgibin/findweather/getForecast?query=ann+arbor
– Model forecasts:
http://www.wunderground.com/modelmaps/maps.asp
?model=NAM&domain=US
Derivation of viscous force
Return to lecture body.
Viscous force (1)
• There is in a fluid friction, which resists the
flow. It is dissipative, and if the fluid is not
otherwise forced, It will slow the fluid and
bring it to rest. Away from boundaries in
the atmosphere this frictional force is often
small, and it is often ignored. (We will
revisit this as we learn more.)
• Close to the boundaries, however, we
have to consider friction.
Return to lecture body.
Viscous force (2)
velocity ≡ u
m/sec
Ocean
Land
Biosphere
velocity must be 0 at surface
Viscous force (3)
velocity ≡ u
m/sec
Ocean
Land
Biosphere
Velocity is zero at the surface; hence,
there is some velocity profile.
Viscous force (4)
(How do we think about this?)
Moving plate with velocity u0
u(h) = u0
h
u(z)
u(z) = (u0-u(0))/h × z
u(0) = 0
Linear velocity profile
Viscous force (5)
(How do we think about this?)
The drag on the moving plate is the same of as the force required
to keep the plate moving. It is proportional to the area (A),
proportional to the velocity of the plate, and inversely proportional to
the distance between the plates; hence,
u(h) = u0
h
u(z)
F = μAu0/h
u(0) = 0
Proportional usually means we assume linear relationship. This is a
model based on observation, and it is an approximation. The constant of
proportionality assumes some physical units. What are they?
Viscous force (6)
(How do we think about this?)
Recognize the u0/h can be represented by ∂u/∂z
u(h) = u0
h
u(z)
F = μA(∂u/∂z)
u(0) = 0
Force per unit area is F/A is defined as shearing stress (t). Like
pressure the shearing stress is proportional to area.
Viscous force (7)
(How do we think about this?)
tzx = μ(∂u/∂z), which is the viscous force
per unit area in the x direction, due to the
variation of velocity in the z direction
u(h) = u0
h
u(z)
Fzx/A = μ(∂u/∂z) ≡ tzx
u(0) = 0
Force per unit area; hence, like pressure.
Viscous force (8)
(Do the same thing we did for pressure)
tzx = tzx0 + (∂tzx/∂z)Dz/2
(x0, y0, z0)
Dz
.
Dy
Dx
tzx0 = stress at (x0, y0, z0)
tzx = -(tzx0 - (∂tzx/∂z)Dz/2)
Viscous force (9)
C
FCzx = (tzx0 + (∂tzx/∂z)Dz/2)DyDx
Dz
.
Dy
D
Dx
FDzx = -(tzx0 - (∂tzx/∂z)Dz/2)DyDx
Viscous force (10)
Fzx = FCzx + FDzx
= (∂tzx/∂z)DzDyDx
We want force per unit mass
Fzx/m = 1/r ∂tzx/∂z
Viscous force (11)
(using definition of t)
Fzx 1 
u

(m )
m r z
z
Viscous force (12)
Assume μ constant
Fzx m  u
u

 2
2
m r z
z
2
2
 ≡ m/r  kinematic viscosity coefficient
Viscous force (13)
Do same for other directions of shear
(variation of velocity)
Frx
u u u
 ( 2  2  2 )
m
x
y
z
2
2
2
Viscous force (14)
Do same for other directions of force
 v  v  v
 ( 2  2  2 )
m
x
y
z
Fry
2
2
2
Frz
 w  w  w
 ( 2  2  2 )
m
x
y
z
2
velocity vector ≡ u = (u,v,w)
2
2
Viscous force (15)
F
2
  (u)
m
Where Laplacian is operating on velocity
vector ≡ u = (u, v, w)
Return to lecture body.
Summary
• Pressure gradient force and viscous force
are examples of surface forces.
• They were proportional to the area of the
surface of our particle of atmosphere.
• They are independent of the mass of the
particle of atmosphere.
• They depend on characteristics of the
particle of atmosphere; characteristics of
the flow.
Return to lecture body.
Our surface forces
du
1
2
  p  (u) 
dt
r
other forces
Now using the text’s convention that the velocity is
u = (u, v, w).
Return to lecture body.
Highs and Lows
Motion initiated by pressure gradient
Opposed by viscosity
Return to lecture body.