AOSS 321, Winter 2009 Earth System Dynamics

Lecture 3 1/15/2009

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• Homework #1 posted today, due on Thursday

(1/22) in class• Our current grader is Kevin Reed (

kareed@umich.edu)• 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, densityThe 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:

€

ρ =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 = ρ Rd T

where Rd is the gas constant for 1 kg of dry air

ρ 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

Mi (g/mol) Volume ratio

N2 28.016 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

Molecular mass of airGasi

€

Σ=28.97 g

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 = Σ partial pressures = Σ pi

• Partial pressure of each gas with index i is:

• Using €

piV =mi

M i

R*T = mi

R*

M i

T

⇔ pi =mi

V

R*

M i

T = ρ iRiT with Ri =R*

M i

€

p = pi

i=1

n

∑ , ρ = ρ i

i=1

n

∑ =mi

Vi=1

n

∑ , m = mi

i=1

n

∑

Ideal gas law: Conversions• It follows:

€

p = pi

i=1

n

∑ =1

V

miR*

M i

T =i=1

n

∑mi

i=1

n

∑ ⎧ ⎨ ⎩

⎫ ⎬ ⎭

V

mi

M i

mi

i=1

n

∑ ⎧ ⎨ ⎩

⎫ ⎬ ⎭

R*Ti=1

n

∑

= ρ

mi

M i

mi

i=1

n

∑

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

R*T =i=1

n

∑ ρTR*

mi

M ii=1

n

∑

mi

i=1

n

∑

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

= ρTR*

Md

= ρTRd

€

=ρ

Conversions• with Md: apparent molecular weight of dry air

• Gas constant for dry air:

€

Md =

mi

i=1

n

∑mi

M ii=1

n

∑= 28.97

g

mol

€

Rd =R*

Md

=8.3145 J mol

28.97 g mol K= 0.287

J

g K

= 287 J

kg K= 287

N m

kg K

Ideal gas law (used in atmospheric dynamics)

• Ideal gas law for dry air:

p = ρ Rd T

• p: pressure in Pa

ρ: 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:

• Pressure (per 1 m2) at point 2:

• Compute pressure differencebetween layer 1 and 2

•

• Rearrange:

p2

p1

z2

z11

2z

Column of air withdensity ρ higher z

higher pz = 0m

€

p = p1 − p2 = ρgz1 − ρgz2

= ρgz1 − ρg z1 + Δz( )

= ρgz1 − ρgz1 − ρgΔz = −ρgΔz

€

p1 = ρgz1

€

p2 = ρgz2 = ρg z1 + Δz( ) with

Δz = z2 − z1

€

p = −ρgΔz ⇔

€

p

Δz= −ρg

Hydrostatic equation

• p: pressure in Pa• z: height in m ρ: density in kg/m3

• g: gravitational acceleration of the Earth, approximately constant, g = 9.81 m/s2

€

∂p

∂z= −ρg

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 ρ is given byF/V = ρ g

Pressure and mass• The atmospheric pressure due to the weight

(per unit area) of the air in an overlying column

• 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:€

p(z) = ρz

∞

∫ gdz'= m(z)g0

€

m(z) = ρz

∞

∫ dz'

Surface Pressure ps

• At the surface

• With m:

vertically integrated mass per unit area (!) of an overlying column of air at the surface, units:kg/m2

• We get:

€

ps = ρ0

∞

∫ gdz'

€

m = ρ0

∞

∫ dz'

€

ps = mg0

1st Exercise

• Compute the pressure profile p(z) with respect to the height z for an isothermal (constant) atmosphere

• Assume a flat Earth (no mountains): The surface height is z0 = 0 m

• Assume the surface pressure p0 = p(z0) is known

€

∂p

∂z= −ρg

€

T

Scale Height

• : 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):

€

H =Rd T

g

€

T

€

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

• Ideal gas law for dry air:

• Hydrostatic equation

• Define the geopotential:

• Note that

Some definitions

€

∂p

∂z= −ρg

€

p = ρRdT

€

Φ= gdz0

z

∫

€

dΦ

dz= g ⇔ dΦ = gdz

Hypsometric equation: Derivation• Express the hydrostatic equation in terms of the

geopotential:

• The variation of the geopotential with respect to pressure only depends on the temperature

€

∂p

∂z= −ρg

⇔∂p

∂z= −

pg

RdT

⇔ −RdT

pdp = gdz = dΦ

⇔ dΦ = −RdTdp

p= −RdT d ln p( )

Hypsometric equation: Derivation

• Integration in the vertical gives:

• Hypsometric equation:

€

Φ(z2) − Φ(z1) = g(Z2 − Z1) = Rd

T

pdp

p2

p1

∫ = Rd Td ln pp2

p1

∫€

dΦz1

z2

∫ = − RdTd(ln p)p1

p2

∫

⇔ Φ(z2) − Φ(z1) = Rd Td(ln p)p21

p1

∫

Hypsometric equation

• With the definition of the geopotential height:

• the hypsometric equation becomes:

• ZT: Thickness of an atmospheric layer (in m) between the pressure surfaces p2 and p1€

ZT = (Z2 − Z1) =Rd

g

T

pdp

p2

p1

∫ =Rd

gTd ln p

p2

p1

∫€

Z =Φ(z)

g(with constant g)

Hypsometric equation for isothermal atmospheres

• If the atmosphere is isothermal with T=T0 the hypsometric equation becomes:

• 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

€

ZT =RdT0

gd ln p

p2

p1

∫ = H lnp1

p2

⎛

⎝ ⎜

⎞

⎠ ⎟

Layer mean temperature

• In an isothermal atmosphere with T0 the layer mean temperature is

• More general definition of the layer mean temperature (for non-isothermal atmospheres):

€

T = T0

€

T =

Td ln pp2

p1

∫

d ln pp2

p1

∫

Examples

• Unit 3 (online resource), frame 20: Dependence on the 800 hPa height on temperaturehttp://www.atmos.washington.edu/2005Q1/101/CD/MAIN3.swf

• Assess the height of the800 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

• Pressure is the same inside and outside at the bottom of the balloon

• How does the pressure at point O compare to the pressure at point I ?

O I

Example: Weather maps• 500 hPa geopotential height map• Blue: isolines that connect equal geopotential

height values (in dm= 10 m)

Weather maps:http://www.rap.ucar.edu/weather/model/

564

576

552

Real weather situations:850 hPa temperature and geopot. thickness

warm air,high geopotentialheight values

cold air, low geopotentialheightvalues

warm

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 (exp (x)) = x,

• exp (ln(x)) = x• xa * xb = xa+b

• (xa)b = xa*b

• (x y)a = xa ya

€

limx →∞

ln x( ) = ∞,limx →0

ln x( ) = −∞€

Integrals• Integrals without limits, a ≠0 is a constant:

€

1

x∫ dx = ln x

€

1

ax∫ dx =

1

aln ax( )

€

exp(x)∫ dx = exp x( )

€

exp(ax)∫ dx =1

aexp ax( )

€

ax n∫ dx =a

n +1( )x n +1 (for n ≠ -1)

€

sin(ax)∫ dx = −1

acos(ax)

€

cos(ax)∫ dx =1

asin(ax)

Derivatives, partial derivatives• Derivatives, a ≠0 is a constant:

• Partial derivatives: treat the variables that arenot differentiated as constants, e.g. f(x,y,z)=xyz

€

d

dxln x =

1

x

€

d

dxexp ax( ) = aexp(ax)

€

d

dxa(x n ) = anx n−1 €

d

dxcos(ax) = −asin(ax)

€

d

dxsin(ax) = acos(ax)

€

∂∂x

(xyz) = yz,∂

∂y(xyz) = xz,

∂

∂z(xyz) = xy

Cartesian Coordinates

xy

z

ij

k

east

north

Local vertical

Velocity vector v = (ui + vj + wk)i: unit vector in x directionj: unit vector in y directonk: unit vector in z direction

Vector notationDefine two vectors, A and B, in a Cartesian coordinate system:

€

A =r A = Axi + Ay j + Azk

A=(Ax,Ay,Az)T is a vector

other symbol: A=(Ax,Ay,Az), also

i, j, k are unit vectors (here in Cartesian coordinates), normalized, orthogonal

€

A = A

€

B =r B = Bxi + By j + Bzk

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:• Vector product:• Gradient of a scalar, e.g. T:• Divergence:• Curl:

€

A • B

€

∇• A

€

A × B

€

∇T

€

∇×A

Unit vectors

Unit vectors in Cartesian coordinates

Unit vectors are normalized: |i| = |j| = |k| = 1

i, j, k are orthogonal:

Example: i•i = 1, i•j = 0, i•k = 0

€

i =

1

0

0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

€

j =

0

1

0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

€

k =

0

0

1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Magnitude and dot product(Cartesian coordinates)

• Magnitude of a vector: |A|

• Scalar (dot) product: A•B = |A| |B| cos() : angle between A and B €

A = Ax2 + Ay

2 + Az2

[ ]

€

A• B = AxBx + AyBy + A zBz

Sum/Subtraction of two vectors(Cartesian coordinates)

• Sum of two vectors: A + B = (Ax+Bx)i +(Ay+By)j +(Az+Bz)k

• Subtraction of two vectors: A - B = (Ax-Bx)i +(Ay-By)j +(Az-Bz)k

Subtraction:AB

Sum:

A

B

-B

Vector (cross) product(Cartesian coordinates)

• Vector product:

• In component form: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)

C

A

B

€

rA ×

r B =

i j k

Ax Ay Az

Bx By Bz

€

A × B = A B sinα

determinant

Gradient(Cartesian coordinates)

• Gradient of a scalar, e.g. temperature T:

• Resulting vector quantity, use of partial derivatives€

∇T =

∂

∂x∂

∂y∂

∂z

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

T =

∂T

∂x∂T

∂y∂T

∂z

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

Horizontal temperature gradients

• Horizontal temperature gradient vector:

• Imagine the isotherms (solid lines)are oriented in the E-W direction

• Draw the horizontal temperature gradient vector.

€

∇hT =

∂T

∂x∂T

∂y

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

cold

warm X

y

Horizontal temperature gradients• Solid lines are isotherms

• Compute the components of the

2D temperature gradient vector:

€

∇hT =

∂T

∂x∂T

∂y

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟≈

ΔT

ΔxΔT

Δy

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

X

yy=100 km

300 K

280 K

270 K

290 Kx=100 km

Horizontal temperature gradients• How does the situation change below?

• Compute the components of the 2D temperature

gradient vector:

X

yy=100 km

300 K

280 K

270 K

290 K

x=100 km

Horizontal temperature gradients

• Draw the horizontal temperature gradient vector.• Solid lines are isotherms, they are 10 K apart.

cold

warm

X

y

Can you compute the components of the gradient vector?

y=100 km

x=100 km