Transcript AOSS 321, Fall 2006 Earth Systems Dynamics 10/9/2006

AOSS 321, Winter 2009
Earth System Dynamics
Lecture 3
1/15/2009
Christiane Jablonowski
[email protected]
734-763-6238
Eric Hetland
[email protected]
734-615-3177
Class News
• Class web site:
https://ctools.umich.edu/portal
• Homework #1 posted today, due on Thursday
(1/22) in class
• Our current grader is Kevin Reed
([email protected])
• Office Hours
– Easiest: contact us after the lectures
– Prof. Jablonowski, 1541B SRB: Tuesday after
class 12:30-1:30pm, Wednesday 4:30-5:30pm
– Prof. Hetland, 2534 C.C. Little, office hour TBA
Today’s lecture
• Ideal gas law for dry air
• Pressure
• Hydrostatic equation
• Scale height
• Geopotential, geopotential height
• Hypsometric equation: Thickness
• Layer-mean temperature
• Mathematical tools
Pressure, temperature, density
• Unit 3: Pressure (online resource)
http://www.atmos.washington.edu/2005Q1/101/
CD/MAIN3.swf
• Consider a hot air balloon
• What is the pressure difference
between the inside and outside
of the hot air balloon?
• What is the temperature difference
between the inside and outside
of the balloon?
• What is the density difference?
Pressure, temperature, density
The ideal gas law
• Pressure is proportional to temperature times
density
• Temperature is proportional to the average
kinetic energy (mass times velocity squared) of
the molecules of a gas
• Density is the mass divided by the volume:
m

V
Ideal gas law
• One form of the ideal gas law is:
p V = n R* T
where the number of moles n in mass m (in
grams) of a substance is given by
n=m/M
• M is the molecular weight of a substance,
e.g. for nitrogen (N2) in the gas phase:
M = 28.0116 g / mol
• p, V, T: pressure, volume and temperature
• R* = 8.3145 J/(K mol): universal gas constant
Another form of the ideal gas law
• Another form of the ideal gas law for dry air:
p =  Rd T
where Rd is the gas constant for 1 kg of dry air
•  is the dry air density, units are kg/m3
• Dry air is a mix of different gases, mainly nitrogen
N2, oxygen O2, argon (Ar), carbon dioxide CO2
Gasi
N2
Mi (g/mol)
28.016
Volume ratio Molecular mass of air
0.7809
21.88
O2
32.000
0.2095
6.704
Ar
39.444
0.0093
0.373
CO2
44.010
0.0003
0.013
g
  28.97
mol
Ideal gas law: Conversions
• Dalton’s law applies:
– each gas completely occupies the whole volume
– each gas obeys its own pressure law
– total dry air pressure =  partial pressures =  pi
• Partial pressure of each gas with index i is:
mi *
R*
piV 
R T  mi
T
Mi
Mi
m i R*
 pi 
T  i RiT
V Mi
n
n
R*
with Ri 
Mi
n
n
mi
• Using p   pi ,    i   , m   mi
i1
i1
i1 V
i1
Ideal gas law: Conversions
• It follows:

 n
 m
i
m



i
n
n
n
1 m i R*
i1  M i
*
p   pi  
T 
R
T
n


V Mi
V
i1
i1
i1


 mi 
i1 
 m 
 n m 
i
i





n
*
 M i  *


M
R
   n
 TRd
R T TR* i1n i  T
Md



i1 
mi
mi




i1 

 i1 


Conversions
• with Md: apparent molecular weight of dry air
n
m
i
g
Md 
 28.97
mol
mi
M
i
i 1
i 1
n
• Gas constant for dry air:
R*
8.3145 J mol
J
Rd 

 0.287
M d 28.97 g mol K
gK
J
Nm
 287
 287
kg K
kg K
Ideal gas law
(used in atmospheric dynamics)
• Ideal gas law for dry air:
p =  Rd T
• p: pressure in Pa
• : density in kg/m3
• Rd: gas constant for 1 kg of dry air
Rd = 287.0 J /(K kg)
• T: temperature in K
Hydrostatic Equation: Derivation
• Pressure (per 1 m2) at point 1:
p1  gz1
• Pressure (per 1 m2) at point 2:
p2  gz2  gz1  z with
z  z2  z1
• Compute pressure difference
between layer 1 and 2
• p  p1  p2  gz1  gz2
Column of
air with
density 
2
p2 z2
1
p1 z1
• Rearrange: p   gz 
z
z = 0m
higher p
 gz1  gz1  z
 gz1  gz1  gz   gz
higher z
p
  g
z
Hydrostatic equation
p
  g
z
•
•
•
•
p: pressure in Pa
z: height in m
3
: density
in
kg/m

g: gravitational acceleration of the Earth,
approximately constant, g = 9.81 m/s2
Pressure
• Pressure decreases with height in the atmosphere
– Why?
• Pressure increases with depth in the ocean
– Why?
• Atmosphere exerts a downward force on the
underlying surface due to the Earth’s gravitational
acceleration
• Downward force (here: the weight force) of a unit
volume (1 m3) of air with density  is given by
F/V =  g
Pressure and mass
• The atmospheric pressure due to the weight
(per unit area) of the air in an overlying column
p(z) 

 gdz' m(z)g
0
z
• Usually we assume that the gravitational
acceleration g is constant: g = g0
• m: vertically integrated mass per unit area of an

overlying column of air at height z:
m(z) 

 dz'
z
Surface Pressure ps
• At the surface
ps 

 gdz'
0
• With m:
m

 dz'
0
 integrated mass per unit area (!) of an
vertically
overlying column of air at the surface, units:kg/m2

• We get:
ps  mg 0
1st Exercise
p
  g
z
• Compute the pressure profile p(z) with respect
to the height z for an isothermal (constant)
atmosphere T
 a flat Earth (no mountains): The
• Assume
surface height is z0 = 0 m
• Assume the surface pressure p0 = p(z0) is

known
Scale Height
Rd T
H
g
• T : average (constant) temperature
• Scale height of the atmosphere: around 8km
• This is the height where the pressure of the
atmosphere is reduced by a factor of 1/e

• For isothermal atmosphere with z0 = 0 m (as seen
before):
p  p0 exp( z /H)
2nd Exercise
• Compute the pressure profile p(z) with respect
to the height z for an atmosphere whose
temperature varies with height like:
T = T0 -  z
• T0 = T(z0) is the known surface temperature at
the surface height z0
•  is a known, constant vertical temperature
gradient (lapse rate)  = -∂T/∂z, units K/m
• Assume a flat Earth (no mountains): z0 = 0 m
• Assume the surface pressure p0 = p(z0) is
known
Some definitions
• Ideal gas law for dry air:
p  Rd T
• Hydrostatic equation

p
  g
z
• Define the geopotential:  
d
 g  d  gdz
• Note
 that
dz
z
 gdz
0
Hypsometric equation: Derivation
• Express the hydrostatic equation in terms of the
geopotential: p
  g
z
p
pg
 
z
Rd T
Rd T

dp  gdz  d
p
dp
 d  Rd T
 Rd T d ln p
p
• The variation of the geopotential with respect to
pressure only depends on the temperature
Hypsometric equation: Derivation
• Integration in the vertical gives:
z2
p2
 d    R Td(ln p)
d
z1
p1
 (z2 )  (z1 )  Rd
p1
 Td(ln p)
p 21
• Hypsometric equation:
p1

p1
T
(z2 )  (z1)  g(Z2  Z1)  Rd  dp  Rd  Tdln p
p2 p
p2
Hypsometric equation
• With the definition of the geopotential height:
(z)
(with constant g)
Z
g
• the hypsometric equation becomes:
Rd

ZT  (Z2  Z1) 
g

p1
T
Rd
 p dp  g
p2
p1
 Tdln p
p2
• ZT: Thickness of an atmospheric layer (in m)
between the pressure surfaces p2 and p1
Hypsometric equation for
isothermal atmospheres
• If the atmosphere is isothermal with T=T0
the hypsometric equation becomes:
Rd T0
ZT 
g
 p1 
 dln p  H lnp 
2
p2
p1
• with scale height H = RdT0/g, g = 9.81m s-2
• We see: the thickness of a layer bounded by

two isobaric pressure surfaces is proportional
to the mean temperature (here constant T0) of
the layer
Layer mean temperature
• In an isothermal atmosphere with T0 the layer
mean temperature is
T  T0
• More general definition of the layer mean
temperature (for non-isothermal atmospheres):
p1

 Tdln p
T
p2
p1
 d ln p
p2
Examples
• Unit 3 (online resource), frame 20: Dependence
on the 800 hPa height on temperature
http://www.atmos.washington.edu/2005Q1/101/C
D/MAIN3.swf
• Assess the height of the
800 hPa surface on a
cold and hot day:
Source: Dale Durran,
University of Washington
Hot air balloon revisited
• Pressure is the same at
the top of the two columns.
hotter (red) column is
taller than cold (blue)
column
O
I
• How does the pressure at
point O compare to the
pressure at point I ?
• Pressure is the same
inside and outside at
the bottom of the balloon
Example: Weather maps
• 500 hPa geopotential height map
• Blue: isolines that connect equal geopotential
height values (in dm= 10 m)
552
564
576
Weather maps:http://www.rap.ucar.edu/weather/model/
Real weather situations:
850 hPa temperature and geopot. thickness
warm
cold air, low
geopotential
height
values
warm air,
high
geopotential
height
values
cold
Mathematical tools: We use
• Logarithms and exponential functions
• Integrals
• Derivatives:
– Partial derivatives
– Chain rule, product rule
• Sine, cosine
• Vectors and vector calculus
• Operators:
– Gradient
– Divergence
– Vector (cross) product
• Spherical coordinates
Logarithms and exponentials
• Natural logarithm, manipulations
–
–
–
–
–
–
•
•
•
•
ln (x) for x > 0
ln (x/y) = ln(x) - ln(y)
ln (x*y) = ln(x) + ln (y)
a * ln(x) = ln (xa)
Values: ln(1) = 0,

ln x  ,limln x  

ln (exp (x)) = x, lim
x
x0
exp (ln(x)) = x
xa * xb = xa+b

(xa)b = xa*b
(x y)a = xa ya
Integrals
• Integrals without limits, a ≠0 is a constant:
1
 x dx  ln x
1
1
 axdx  a lnax
1
 sin(ax)dx   a cos(ax)
1
 cos(ax)dx  a sin(ax)


 exp(x)dx  expx

1 
 exp(ax)dx  a expax
a
n
n 1
ax
dx

x
(for n -1)

n  1

Derivatives, partial derivatives
• Derivatives, a ≠0 is a constant:
d
1
d
ln x 
cos(ax)  asin(ax)
dx
x
dx
d
d
expax  aexp(ax)
sin(ax)  acos(ax)
dx
dx
d

a(x n ) = anxn1 
dx
 • Partial derivatives: 
treat the variables that are

not differentiated as constants, e.g. f(x,y,z)=xyz



(xyz )  yz, (xyz )  xz, (xyz )  xy
x
y
z
Cartesian Coordinates
Local
vertical
z k
north
y j
x i
east
Velocity vector v = (ui + vj + wk)
i: unit vector in x direction
j: unit vector in y directon
k: unit vector in z direction
Vector notation
Define two vectors, A and B, in a Cartesian
coordinate system:
A  A  Ax i  Ay j Azk
B = B  Bx i  By j  Bzk


A=(Ax,Ay,Az)T is a vector
other symbol: A=(Ax,Ay,Az), also
AA
i, j, k are unit vectors (here in Cartesian
coordinates), normalized, orthogonal

Vector calculus
Are you comfortable with:
• c * A, c is a scalar constant
• Magnitude of a vector A: |A|
• A + B, A - B
• Scalar product of two vectors: A B
• Vector product: A B
• Gradient of a scalar, e.g. T: T
• Divergence:  A 
• Curl: A


Unit vectors
Unit vectors in Cartesian coordinates
1
 
i  0
 
0
0
 
j  1
 
0
0
 
k  0
 
1
Unit vectors are normalized: |i| = |j| = |k| = 1



i, j, k are orthogonal:
Example: i•i = 1, i•j = 0, i•k = 0
Magnitude and dot product
(Cartesian coordinates)
• Magnitude of a vector: |A|
A
2
2
2
A

A

A
 x y z
• Scalar (dot) product: A•B = |A| |B| cos()
 angle between A and B
:
A B  Ax Bx + AyBy + AzBz
Sum/Subtraction of two vectors
(Cartesian coordinates)
• Sum of two vectors:
A + B = (Ax+Bx)i +(Ay+By)j +(Az+Bz)k
Sum:
B
A
• Subtraction of two vectors:
A - B = (Ax-Bx)i +(Ay-By)j +(Az-Bz)k
B
Subtraction:
-B
A

Vector (cross) product
(Cartesian coordinates)
• Vector product: A  B  A B sin 
i
A  B = Ax

Bx
j
k
Ay
By
Az
Bz
• In component form:
C
B

A
determinant
C = A x B = (AyBz - AzBy)i + (AzBx - AxBz)j + (AxBy - AyBx)k
• The resultant vector C is in the plane that is
perpendicular to the plane that contains A and B
(direction: right hand rule)
Gradient
(Cartesian coordinates)
• Gradient of a scalar, e.g. temperature T:
   T 
   
x  x 
  T 

T =
T
y  y 
   T 
   
z  z 
• Resulting vector quantity, use of partial derivatives

Horizontal temperature gradients
T 
x 
• Horizontal temperature gradient vector:  h T  T 
 
y 
• Imagine the isotherms (solid lines)are oriented
in the E-W direction
cold

y
warm
X
• Draw the horizontal temperature gradient vector.
Horizontal temperature gradients
• Solid lines are isotherms
T  T 
• Compute the components of the
x  x 
2D temperature gradient vector:  h T     
T
T
   
y  y 
270 K
y=100 km
280 K

y
290 K
x=100 km
X
300 K
Horizontal temperature gradients
• How does the situation change below?
• Compute the components of the 2D temperature
gradient vector:
270 K
y=100 km
280 K
y
290 K
X
x=100 km
300 K
Horizontal temperature gradients
• Draw the horizontal temperature gradient vector.
• Solid lines are isotherms, they are 10 K apart.
y
Can you compute
the components of
the gradient vector?
X
y=100 km
x=100 km