Transcript Topic # 3

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   Rd T
p = pressure;  = density; Rd = gas constant (???);
T = temperature.
“Used” in virtually everything – i.e., it is almost
always assumed!
2.
Application: The Hydrostatic Equation
p
  g
z
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2.
Also –
geopotential
geopotential height
thickness
3.
Thermo: The First Law of Thermodynamics
Also:
specific heat
enthalpy (?)
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3.
Applications:
adiabatic processes
dry adiabatic lapse rate
potential temperature
thermodynamic diagrams
4.
Water Vapor in the air
Measures of vapor in the air
mixing ratio
specific humidity
relative humidity
dew 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
entropy
Clausius-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:
pV  mRT
(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)
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Value of R depends on the gas involved – see below.
We also write:
p   RT
where  = density (kg/m3), or:
p  RT
where  is specific volume = 1/.
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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,
pd  d  Rd T or pd   d Rd T
where Rd = gas constant for dry air.
Value/meaning of Rd?
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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 grams
water has M = 18.015 – one mol of water weighs 18.015 grams
CO2 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=nM
Thus:
 R* 
pV  m T
M 
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!
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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:
pd  d  Rd T or pd   d Rd T
Rd = R*/ Md = 287 J K-1 kg-1
For moist air???
Look at pure water first – then combine.
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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
ev = RvT
(3.12)
where e = pressure due to water vapor alone
v is the specific volume of the vapor
and 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:
Tv 
T
e
1  (1   )
p
 = Rd/Rv = 0.622
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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!
p, , T
pe, e, Te
and vice versa if p < pd
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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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p
  g
z
The hydrostatic equation
(remember it!)
And of course at the same
time:
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p   Rd 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
z+dz
z
done = force x distance work
= (mass x acceleration) x distance
= (1 x g) x dz
= gdz
We define geopotential, , by: d = gdz
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d = gdz
And if we treat g as constant and add up over z
(“integrate”), we get:
 = gz
Units? m2s-2 – energy units
To make these units and values more meaningful, we
define:
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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
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energy units
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Z=5822 gpm
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and z5822 m !!
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We can also talk about thickness…
z2
z1
Thickness of a
layer
Thickness = Z2 - Z1
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Now we recall:
p
  g
z
So:
Rd Tv
gdz  d  dp  
dp
p
So by integration:
Rd p2 Tv
Z 2  Z1  
dp

9.8 p1 p
(thickness)
And we need values of Tv to get any further…
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Suppose Tv is constant ( Tv ) in a layer (isothermal).
Then:
Rd p2 Tv
Rd Tv  p2  Rd Tv  p1 
Z 2  Z1  
dp  
ln 
ln 



p
p
p
1
2
9.8 1 p
9.8 
9.8 
Two things:
a)
Thickness  (layer mean) temperature – obvious!
b)
If we set:
Rd Tv Rd T
H

9.8
g
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Then:
Z 2  Z1  H ln  p1 
 p2 
Rearranging:


Z

Z


2
1
p2  p1 exp 

H


This is basically the same as Eq.(1.8)!!!
As before, H is scale height, and now
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Rd T
H
g
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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:
Z 2  Z1 
Rd Tv  p1 
ln 

 p2 
go
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.
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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:
du = dq – dw
This defines du
and we will show that du  T
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-add dq of energy
-parcel may expand (dw)
-internal temperature may change (du, dT)
1) add dq
2) possible expansion
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3) possible T change
internally
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So…
du = dq – dw
Write more usefully…
For us, the main work (dw) is expansion/contraction work:
Use:
1.
2.
3.
work = force x distance
Pressure = force per unit area
Assume unit mass where mass = density x volume
To show that:
dw = pd
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Thus…
dq = du + pd
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
Water
Air
Soil
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We expect T(substance) to increase
How much?
We can define specific heat as:
heat added
dq
dT
temp change
More carefully:
 dq 
 dq 
cv  
 , cp  

 dT v
 dT  p
constant volume
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constant pressure
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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…
du = cvdT
dq = cvdT + pd
(3.41)
…another statement of the 1st Law (close to useful)
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Again…
dq = du + pd
Also…
cp = (dq/dT)p
Now…
p = RdT
 d(p) = d(RdT)
= Rdd(T)
Also
d(p) = pd + dp
(Gas Law)
(Rd constant!)
(math)
Rearranging…
pd = d(p) - dp = RddT - dp
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So…
dq = du + pd
dq = du +RddT - dp
dq = cvdT + RddT - dp
dq = (cv + Rd)dT - dp
(previous slide)
(2 slides back)
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
or
(1004 = 717+287 ???)(yes!!!)
Rd = cp - cv
And
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dq = cpdT - dp
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(3.46)
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dq = cpdT - dp
Probably the most useful form of the 1st Law…
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Final note…
dh = cpdT
defines enthalpy, h
Next useful concept…adiabatic processes
a)
b)
What does “adiabatic” mean?
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…
So…

dq=0
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,
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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…
-dT/dz = g/cp  d
…the dry adiabatic lapse rate (dalr, 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…
 po 
  T  
 p
Rd
cp
Notes:
•
•
•
•
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 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  
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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
z
20 km
10 km
T
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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
lnp
20 km
10 km
T
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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…
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