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MET 60

Chapter 3:

Atmospheric Thermodynamics

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In chapters 3, 4, and 6, we will be zooming in and out from the largest-scale to the molecular/atomic scale (ditto chapter 5 but…)

http://www.youtube.com/watch?v=Sfpb9GqYLiI Ditto chapter 5 but …

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The layout of chapter 3 is:

Some thermodynamics (“thermo” or “TD”)

• Application(s)

More thermo

• More applications

Yet more thermo

• Yet more applications

Etc.

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1. Basic Thermo: The Ideal Gas Law (or the Equation of State)

p = pressure; = density; Rd = gas constant (???); T = temperature.

“Used” in virtually everything – i.e., it is almost always assumed!

2. Application: The Hydrostatic Equation

dp R T

p gz

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2. Also –

geopotentialgeopotential heightthickness

3. Thermo: The First Law of Thermodynamics

Also:

specific heatenthalpy (?)

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3. Applications:

adiabatic processesdry adiabatic lapse ratepotential temperaturethermodynamic diagrams

4. Water Vapor in the air

Measures of vapor in the airmixing ratiospecific humidityrelative humiditydew point

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4. Water Vapor in the air (cont.)

lifting condensation level (LCL)latent heat

Chinook winds – moist ascent followed by dry descent

5. Application: Static stability

As a parcel of air is lifted (or rises), what can happen to it?

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If T(parcel) > T(environment), parcel rises (A)

If T(parcel) < T(environment), parcel sinks (B)

(A) Is termed an unstable situation, and air parcels can rise deep clouds (e.g., Cb)

(B) Is termed stable situation, no deep clouds BUT gravity waves in cloud patterns

Fig. 3.14

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6. Thermo: The Second Law of Thermo

entropyClausius-Clapeyron Equation

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The Ideal Gas Law (or the Equation of State)

It is experimentally found that all gases obey this relation:

(3.1)

p = pressure (Pa – not mb or hPa)V = volume (m3)m = mass (kg)T = temperature (degrees K – not C or F)R is a constant of proportionality (the gas constant)

pV mRT

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Value of R depends on the gas involved – see below.

We also write:

where = density (kg/m3), or:

where is specific volume = 1/.

p RT

p RT

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Two special cases (which we often look at!):

1) Constant temperature… isothermal …

Volume is inversely proportional to _____________

This is Boyle’s Law (1600’s !!)

2) Constant pressure…

Volume is proportional to _____________

This is Charles’ Law (late 1700’s – early 1800’s)

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3) Constant volume…

Pressure is proportional to _____________

This is also Charles’ Law

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For dry air,

where Rd = gas constant for dry air.

Value/meaning of Rd?

d d d d d dp R T or p R T

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Consider any (pure) gas…

a) It has a molecular weight = M

b) A mole (or mol) of this substance is defined as the number value of M, expressed in grams

Examples:

oxygen has M = 32 – one mol of O2 weighs 32 gramswater has M = 18.015 – one mol of water weighs 18.015 gramsCO2 has M = 44.01 – one mol of CO2 weighs 44.01 grams

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For all substances, a mol of the substance has the same number of molecules – NA (Avogadro’s number)

Thus in Eq. (3.1), for 1 mol of gas, the constant (R) is the same for all gases.

This is the Universal Gas Constant (R*).

So for 1 mol: pV = R*T (unit mass)

And for n moles: pV = nR*T (mass “n”)

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Now if we have a mass m of the gas, then:

m = n M

Thus:

So that (R*/M) serves as a “gas constant for that species”.

So – for dry air – we just need the value of M – call it Md!

TMRmpV

*

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We know the components of dry air (78% N2, 21% O2 etc.) and we know the molecular weight of each of these.

We compute Md from (3.10) to get

Md = 28.97

And thus,

Rd = R*/ Md = 287 J K-1 kg-1

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Recap…

Ideal Gas Law / Eqn. Of State for dry air:

Rd = R*/ Md = 287 J K-1 kg-1

For moist air???

Look at pure water first – then combine.

d d d d d dp R T or p R T

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For water vapor…

Mw = 18.016

So Rv = R*/Mw = 461.5 J K-1 kg-1

And the ideal gas equation is

ev = RvT (3.12)

where e = pressure due to water vapor alonev is the specific volume of the vaporand Rv is the gas constant for water vapor

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For moist air?

We use Dalton’s Law of partial pressures:

Total pressure = sum of individual pressures

(as long as the gases do not interact chemically!)

Example 3.1

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For moist air, we could use an Rv

but the value would depend on how much moisture is in the air (not constant!)

So Rv would be a “variable constant”!!!

Instead, we introduce:

Virtual Temperature (Tv)

A fictitious temperature dry air would need to have in order to have the same density as moist air @ same pressure

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Note: (moist air ) < (dry air)

Because…

Also…

Tv > T (for moist air)

In practice, T and Tv differ by only a few degrees

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W&H derive:

= Rd/Rv = 0.622

)1(1

peTTv

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The Hydrostatic Equation

Consider a parcel of (dry) air…

p, , T

pe, e, Te

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Suppose p > pd

p, , T

pe, e, Te

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The parcel expands!

and vice versa if p < pd

p, , T

pe, e, Te

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What mattered was the pressure difference or pressure gradient

-not the actual pressure

Now apply this thinking to a layer of air

-Fig. 3.1

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The hydrostatic equation (remember it!)

And of course at the same time:

p gz

dp R T

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Some applications…

using the hydrostatic equation and the equation of state

Look at a 500 mb/hPa map

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The contours link locations where the 500 hPa surface is at the same “height” ASL.

Expressed in decameters (dm or dam)

Example: a height of 5000 m (5 km) is contoured as 500 dm

As we will see,

height(500 hPa) temperature of (1000-500 hPa) layer

thickness

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Geopotential

From the 500 mb chart, consider a parcel on the 576 dm contour.

To get there means that work has been done against the force of gravity.

Geopotential is defined as the work done to raise a mass of 1 kg from altitude z=0 to a desired altitude (z).

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Geopotential

done = force x distance work = (mass x acceleration) x distance = (1 x g) x dz = gdz

We define geopotential, , by: d = gdz

z+dz

z

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d = gdz

And if we treat g as constant and add up over z (“integrate”), we get:

Units? m2s-2 – energy units

To make these units and values more meaningful, we define:

= gz

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Geopotential height:

Z = /9.8 = {gz / 9.8} m2s-2

For the troposphere, Z z.

We even define a geopotential meter (gpm) such that:

at an altitude z meters, the geopotential height is Z gpm, where the numerical values are about the same

Example: at height 5000 meters, Z 5000 gpm

height units energy units

9/24/09 MET 60 topic 03 37Z=5822 gpm

and z5822 m !!

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We can also talk about thickness…

Thickness = Z2 - Z1

z2

z1

Thickness of a layer

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Now we recall:

So:

So by integration:

(thickness)

And we need values of Tv to get any further…

p gz

dppTRdpdgdz vd

2

18.912

p

pvd dppTRZZ

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Suppose Tv is constant ( ) in a layer (isothermal).

Then:

Two things:

a) Thickness (layer mean) temperature – obvious!

b) If we set:

2

1

1

212 ln

8.9ln

8.98.92

1 ppTR

ppTRdp

pTRZZ vdvdp

pvd

gTRTRH dvd

8.9

vT

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Then:

Rearranging:

This is basically the same as Eq.(1.8)!!!

As before, H is scale height, and now

2

112 ln p

pHZZ

H

ZZpp 1212 exp

gTRH d

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What if Tv is NOT constant (which it will not be)?

We can still define a layer-average (Eq. (3.28)) and write:

where go is a constant value of g (9.8).

This is called the hypsometric equation.

-- Use above form to find heights given pressures.-- Invert and use to find pressures given heights.

12 1

2lnd v

o

R T pZ Z pg

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Warm-core lows…

Warm at its center

Example: a hurricane (Fig. (3.3))

Intensity decreases with height

Cold-core lows…

Cold at its center

Example: a mid-latitude cyclone

Intensity increases with height

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The First Law of Thermodynamics

Consider a closed system (e.g., a parcel of air).

It has internal energy (“u”) = energy due to molecular kinetic and potential energies

Suppose some energy (dq) is added to the system

Example: via radiation from the sun

What happens?

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Some of the energy goes into work done (dw) by the system against its surroundings

Example: expansion

What’s left is a change in internal energy:

This defines du

and we will show that du T

du = dq – dw

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-add dq of energy

-parcel may expand (dw)

-internal temperature may change (du, dT)

1) add dq

2) possible expansion

3) possible T change internally

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So…

Write more usefully…

For us, the main work (dw) is expansion/contraction work:

Use: 1. work = force x distance2. Pressure = force per unit area3. Assume unit mass where mass = density x volume

To show that:

dw = pd

du = dq – dw

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Thus…

which is closer to useful (see more below…)

Next useful concept…specific heat

Suppose we add some thermal energy (dq) to a unit mass of a substance

WaterAirSoil

dq = du + pd

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We expect T(substance) to increase

How much?

We can define specific heat as:

More carefully:

constant volume constant pressure

dTdq

pp

vv dT

dqcdTdqc

,

heat added

temp change

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Specific heat is the heat energy needed to raise the temperature of a unit mass of substance by one degree.

Values (p. 467!!)?

Air cp = 1004 J K-1 kg-1

cv = 717 J K-1 kg-1

Water…liquid cw = 4218 J K-1 kg-1

Soil 5x lower than water

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Again… dq = du + pd … expression for du???

And… cv = (dq/dT)v

But…volume constant no expansion work pd = 0 dq = du above cv = (du/dT)v = (du/dT) (see text)

And thus… (3.41)

…another statement of the 1st Law (close to useful)

du = cvdT

dq = cvdT + pd

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Again… dq = du + pd

Also… cp = (dq/dT)p

Now… p = RdT (Gas Law)

d(p) = d(RdT) = Rdd(T) (Rd constant!)

Also d(p) = pd + dp (math)

Rearranging… pd = d(p) - dp = RddT - dp

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So… dq = du + pd

dq = du +RddT - dp (previous slide) dq = cvdT + RddT - dp (2 slides back) dq = (cv + Rd)dT - dp

At constant pressure, last term is zero (dp=0), and also cp = (dq/dT)p dq = cpdT (on LHS)

Putting together cpdT = (cv + Rd)dT cp = cv + Rd(1004 = 717+287 ???)(yes!!!)or

And (3.46)Rd = cp - cv

dq = cpdT - dp

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Probably the most useful form of the 1st Law…

dq = cpdT - dp

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Final note…

dh = cpdT

defines enthalpy, h

Next useful concept…adiabatic processes

a) What does “adiabatic” mean?b) What does “adiabatic” imply for a TD system?

Recall the 1st Law: dq = du + pd

Adiabatic means there is zero heat added/subtracted (physical meaning)

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Or… dq=0

So… dq = cvdT + pd

cvdT + pd = 0 or cpdT - dp = 0

(mathematical implication)

Suppose an air parcel (see p.77) rises adiabatically…

What happens to T(parcel)?

From above, cpdT - dp = 0

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cpdT = dp

BUT…hydrostatic equation gives

dp = - gdz

Thus… cpdT = dp = - gdz

And so the lapse rate (-dT/dz) is…

…the dry adiabatic lapse rate (dalr, d)

-dT/dz = g/cp d

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Value??

g/cp = 9.8 / 1004 10 / 1000 = 10 C/km

Potential temperature

Suppose a parcel is at height z m, pressure level p hPa, and has temperature T

Suppose we bring the parcel to sea level (z=0, p=po) adiabatically.

It would compress and warm to a certain temperature.

We call this the potential temperature, .

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WH show…

Notes:

is hypothetical/fictitious (again!!! Like Tv)• We almost always use po = 1000 hPa is used A LOT (not “alot”)• Rd/cp = 287/1004 = 2/7

pdc

R

o

ppT

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Thermodynamic diagrams

Much more next semester in the lab

For now, some ideas.

Suppose we have data such as temperature versus height.

Suppose we want to plot this.

How?

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Most obvious…T versus z

T

z20 km

10 km

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Problem…we don’t measure z.

We do measure p (e.g., in a radiosonde sounding).

Maybe plot T versus p?

Problem…p falls off exponentially with height.

Solution…plot T versus lnp!!

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T versus ln(p) = emagram

T

lnp20 km

10 km

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Requirement…area on the plot work done during a cyclic TD process (pd)

Emagram has this property!

For easier interpretation, we “skew” the axis

Result: “skew-T lnp” diagram

Background info:

http://www.atmos.millersville.edu/~lead/SkewT_HowTo.html

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Today’s sounding…