9/24/09MET 60 topic 031 MET 60 Chapter 3: Atmospheric Thermodynamics.
-
Upload
kelly-king -
Category
Documents
-
view
240 -
download
0
description
Transcript of 9/24/09MET 60 topic 031 MET 60 Chapter 3: Atmospheric Thermodynamics.
9/24/09 MET 60 topic 03 1
MET 60
Chapter 3:
Atmospheric Thermodynamics
9/24/09 MET 60 topic 03 2
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 …
9/24/09 MET 60 topic 03 3
The layout of chapter 3 is:
Some thermodynamics (“thermo” or “TD”)
• Application(s)
More thermo
• More applications
Yet more thermo
• Yet more applications
Etc.
9/24/09 MET 60 topic 03 4
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
9/24/09 MET 60 topic 03 5
2. Also –
geopotentialgeopotential heightthickness
3. Thermo: The First Law of Thermodynamics
Also:
specific heatenthalpy (?)
9/24/09 MET 60 topic 03 6
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
9/24/09 MET 60 topic 03 7
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?
9/24/09 MET 60 topic 03 8
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
9/24/09 MET 60 topic 03 9
6. Thermo: The Second Law of Thermo
entropyClausius-Clapeyron Equation
9/24/09 MET 60 topic 03 10
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
9/24/09 MET 60 topic 03 11
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
9/24/09 MET 60 topic 03 12
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)
9/24/09 MET 60 topic 03 13
3) Constant volume…
Pressure is proportional to _____________
This is also Charles’ Law
9/24/09 MET 60 topic 03 14
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
9/24/09 MET 60 topic 03 15
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
9/24/09 MET 60 topic 03 16
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”)
9/24/09 MET 60 topic 03 17
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
*
9/24/09 MET 60 topic 03 18
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
9/24/09 MET 60 topic 03 19
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
9/24/09 MET 60 topic 03 20
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
9/24/09 MET 60 topic 03 21
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
9/24/09 MET 60 topic 03 22
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
9/24/09 MET 60 topic 03 23
Note: (moist air ) < (dry air)
Because…
Also…
Tv > T (for moist air)
In practice, T and Tv differ by only a few degrees
9/24/09 MET 60 topic 03 24
W&H derive:
= Rd/Rv = 0.622
)1(1
peTTv
9/24/09 MET 60 topic 03 25
The Hydrostatic Equation
Consider a parcel of (dry) air…
p, , T
pe, e, Te
9/24/09 MET 60 topic 03 26
Suppose p > pd
p, , T
pe, e, Te
9/24/09 MET 60 topic 03 27
The parcel expands!
and vice versa if p < pd
p, , T
pe, e, Te
9/24/09 MET 60 topic 03 28
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
9/24/09 MET 60 topic 03 29
The hydrostatic equation (remember it!)
And of course at the same time:
p gz
dp R T
9/24/09 MET 60 topic 03 30
Some applications…
using the hydrostatic equation and the equation of state
Look at a 500 mb/hPa map
9/24/09 MET 60 topic 03 31
9/24/09 MET 60 topic 03 32
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
9/24/09 MET 60 topic 03 33
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).
9/24/09 MET 60 topic 03 34
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
9/24/09 MET 60 topic 03 35
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
9/24/09 MET 60 topic 03 36
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 !!
9/24/09 MET 60 topic 03 38
We can also talk about thickness…
Thickness = Z2 - Z1
z2
z1
Thickness of a layer
9/24/09 MET 60 topic 03 39
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
9/24/09 MET 60 topic 03 40
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
9/24/09 MET 60 topic 03 41
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
9/24/09 MET 60 topic 03 42
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
9/24/09 MET 60 topic 03 43
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
9/24/09 MET 60 topic 03 44
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?
9/24/09 MET 60 topic 03 45
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
9/24/09 MET 60 topic 03 46
-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
9/24/09 MET 60 topic 03 47
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
9/24/09 MET 60 topic 03 48
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
9/24/09 MET 60 topic 03 49
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
9/24/09 MET 60 topic 03 50
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
9/24/09 MET 60 topic 03 51
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
9/24/09 MET 60 topic 03 52
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
9/24/09 MET 60 topic 03 53
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
9/24/09 MET 60 topic 03 54
Probably the most useful form of the 1st Law…
dq = cpdT - dp
9/24/09 MET 60 topic 03 55
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)
9/24/09 MET 60 topic 03 56
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
9/24/09 MET 60 topic 03 57
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
9/24/09 MET 60 topic 03 58
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, .
9/24/09 MET 60 topic 03 59
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
9/24/09 MET 60 topic 03 60
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?
9/24/09 MET 60 topic 03 61
Most obvious…T versus z
T
z20 km
10 km
9/24/09 MET 60 topic 03 62
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!!
9/24/09 MET 60 topic 03 63
T versus ln(p) = emagram
T
lnp20 km
10 km
9/24/09 MET 60 topic 03 64
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
9/24/09 MET 60 topic 03 65
Today’s sounding…