Transcript Atmospheric Science 4310 / 7310

Atmospheric Science 4310 / 7310
Atmospheric
Thermodynamics - II
By
Anthony R. Lupo
Day 1

The work done by an expanding gas

Let’s draw a piston:
Day 1

 Consider a mass of gas at Pressure P in a
cylinder of Cross section A

Now, Recall from Calc III or Physics:

Work = force x distance or Work = Force dot
distance
W   F  ds

So only forces parallel to the distance travelled do
work!
Day 1

Then,
dW  F  ds
 or 
dW
ds
F
dt
dt
Day 1


But, we know that:
Pressure = Force / Unit Area

So then,

Force = P x Area
Day 1

Total work increment now:
dW   P  Area   ds

Well,

Area x length = Volume

soo………………… A * ds = dVol
Day 1

Then we get the result:

 Work :

Let’s “work” with Work per unit mass:
dW  PdV
dW
dV
P
m
m

 Thus, we can start out with volume of only
one 1 kg of gas!!!
Day 1

 And we know that:

a = Vol/m

then……..

Note the “heavy D”
DW  Pda
Day 1

So….
DW  Pda
 or 
a2
W   Pda
a1

Very Important!!! This is not path
independent, it’s not an exact differential!

 Thus is is easy to see that if a parcel
undergoes expansion, or a2 > a1
Day 1

Parcel does work expanding against
environment!

 But if the parcel undergoes contraction
(a2 < a1)

Environment is doing work on the parcel!
Day 1/2

A derivation more relevant to an air parcel

 Given a spherical parcel of air with radius
R, Surface Area = 4pr2 and the volume is
4/3pr3.
Day 2

 We calculate dW in expanding the parcel
from r to r + dr in an environment where the
pressure is p!

Then,

Work = F x dist = Pressure x Area x distance
Day 2

Well,

Dist = dr

And;

Area = A + dA
Day 2

As with the “expanding piston”

DW = p*A*dr = p * dV

  Recall from Calc II (or is it physics?)

dV = 4p r2 dr
Day 2

Then we’ll deevide by the unit mass again to get:

dw = p da

Now the short comings in this problem are obvious:

1.
(Surf)Area = A + dA = 4 p r2 + 8p r dr
Day 2


2. The parcel expands and p decreases;
so: p = p + dp

So….. the REAL problem is:

dW = (p+dp)(A+dA) dr
Day 2

 Which “foils” out to,

DW = pAdr + dpAdr + pdAdr + dpDAdr

then assume;

Dw = p da + dp da + p dA dr + dp dA dr
Day 2

Which then reduces to (by scale analysis):
 Dw
= p da
Day 2

The First Law of Thermodynamics: A derivation.

 The First Law of Thermodynamics, is really just a
statement of the principle of the conservation of
energy.

 This law can be used as a predictive tool: ie, we
can calculate changes in T, a (r), or pressure for a
parcel.
Day 2

 The first law of thermodynamics is associated with
changes in these quantities whereas the equation of
state relates the quantities themselves, at a given
place at a given time (or at a given place for a steady
state).

Forms of energy relevant to our treatment of the
1st Law

“Energy” is (simply) the capacity to do work!

for our purposes, a system is an air parcel.
Day 2

KE (kinetic energy)  associated with
motion, energy of motion

(Definition: “Kinetic” means the study of
motion w/o regard to the forces that cause it)
1
KE  mass  velocity 2
2
 or 


dKE 1
 mass  2V  dV
dt
2
Day 2

PE (potential energy)  is the energy of
position relevant to some reference level, or
within some potential (e.g., gravitational)
field.
PE  mass  g * height
 or 
dPE
dz
 mass  g 
dt
dt
where
dz d
g 
dt dt
Day 2

**The location of Z = 0 is arbitrary, but it makes good
sense to use sea level.

Internal energy (IE)  Is the energy (KE and PE)
associated with the individual molecules in the
parcel.

**So IE is temperature dependent (use Absolute T)!!!
Day 2

Deriving the First Law: (Hess pp 25 – 26)

Statement of Conservation of Energy:
“Energy of all sorts can neither be created
nor destroyed” (Of course we ignore
relativistic theory E = mc2)

Written in incremental form (here it is!):
Day 2/3

The “book definition”:

“Any increment of energy added to a system is equal
to the algebraic sum of the increments in organized
KE, PE, IE (thermal), work done by the system on it’s
surroundings, and whatever forms of electrical and
magnetic energy may appear”

Dheat = DKE + DPE + DIE + DWork + DE&M
Day 2/3

 We can neglect electrical and magnetic for atmospheric
motions we consider (scale analysis).

** Recall: our system is the air parcel!

 Consider the LHS; “increment of energy added to a system”
can be broken down into “heat added to the system” + work
done on the system by the surroundings”.

We want to simplify, if possible to quantify our physical principle
(the conservation of energy).
Day 3

 Kinetic Energy: We first note that
increments of Kinetic energy are small
compared to changes in internal energy and
increments of work done. For example,
consider a parcel of air changing speeds
from 10 m/s to 50 m/s

DKE = ½ (50 m/s)2 – ½ (10 m/s)2 or 1.2 x
103 m2 sec -2
Day 3

Next, let us consider a change of 20 C for a
parcel of air at a constant volume, a change
of internal energy:

DIE = Cv DT = 717 J K-1 kg-1 x 20 K

or 1.4 x 104 m2 sec-2
Day 3

Then, let’s consider the change in PE at, oh,
say 500 hPa,

So,

DPE = g*Dz = 10 m s-2 * 60 m = 600 m2 s-2
Day 3

Pressure work;

We have dw = p da
and pa = RT

so dW = (RT/a) da
= RT d[ln(a)]

Since T = Const, we can write:


= ln 0.1 * 287.04 J/kg K * 280 K
W = 8.47 x 103 m2 s-2
Day 3

So, we’ve shown that:

Dh/dt = IE + PE + KE - Work done on the system by
the surroundings + work done by the system on the
surroundings+ EE + ME

 We neglected EE and ME by scale analysis, and
we can neglect KE and PE by scale analysis (or do
we)?
Day 3


Well,
DKE = DKE(horiz) + DKE (vert)

***KE vertical is very tiny, by several orders of
magnitude (at least 4)!

Work done on the system by surroundings (W):

-(Work horiz + Wvert)
Day 3

 The Horizontal pressure gradient is responsible
for changes in wind speed.

The pressure gradient is in DKE and Wh (which is
horizontal wind times distance covered)

Then, DKE must cancel with Wh!

How about Wv?
Day 3

 Well, DPE = gdz

and,

Wv = vert. PGF x distance = - (1/r) (dp/dz) (dz)

If hydrostatic balance is assumed to hold then g = (1/r) (dp/dz).
Day 3

Thus, DPE + Wv = gdz – g dz = 0 (and these
cancel)!

 So in the incremental form of the first law we are
left with:

Dh/Dt = IE + work done by the system on the
surroundings (1)

**(note, the “heavy D”)
Day 3/4/5

Then if we

define heat added as dh/dt (dq/dt) or just dh or dq

Internal energy as “du”,

and work done by the system as “dw”, then:

We can quantify first law (this is it)!    ?

Dh = du + Dw or du = dh - dw
Day 5

The First Law of Thermodynamics and Entropy.

What happens when heat is added to parcel?

1st Law: dh = du + dw

 Heat can be added (or subtracted, of course)!
How?

Conduction, convection, radiation, or phase change
(All “Diabatic” heating)
Day 5

Diabatic processes:

Sensible heating  warm to cold by contact

Radiative heating  “Short wave in” or “long wave
out”

Latent Heating  phase changes of water mass
Day 5

 So we can say formally “if a small quantity of
heat* is added to the system (parcel) some goes into
increasing the int. energy of the parcel, and some
gets expended as the parcel does work (dw) on the
surroundings”.

*(Note: heat is NOT temperature!)

 We know the 1st law, however, how do we
express in terms of state thermodynamic variables
(T, a, or P?) so we can obtain useful expressions for
atmospheric calculations?
Day 5

 Unfortunately, there are no formal, or analytical,
mathematical expressions for dh (especially for atmospheric
processes – these typically “parameterized” (which is an
empirical “fudge factor”)).

 Thus, we are reduced to solving as a residual, and in the 1st
Law leave as dh/dt or dq/dt (Q-dot).

Well, we showed that the work done by an expanding gas could
be derived by considering a piston (closed controlled system),
or an air parcel.,
Day 5
Thus, dw = p da.
So the 1st Law is now:
Q  du  pda
Day 5

Internal Energy

 With the pressure work term expressed in terms
of state variables p and a, we must now consider
that internal energy may be dependent on (T and a)
or (T and p) (since p and a can be related)

So: Consider that u = u(T,a) or u = u(T,p)
Day 5

The expression
u
dT u
da
du 

T a const dt a T const dt
Day 5

Joule’s Law; and his experiment conceptualize
this term

Consider an insulated container with two chambers,
one filled with Gas at pressure P = Po, and the other
a vaccuum at P = 0. (connected by valve)

Insulated chamber, thus Q dot = 0. (No surroundings
also, so dw = 0).
Day 5

Draw:
Day 5

Then the valve is opened and gas expands into
vacuum chamber (let ‘er rip!) P = 0. (No temp
change occurs), then from first law:

Hmm, Dq = du + dw

Well:

1. Dq = 0 (no heat added)
2. dw = 0 (since chamber cannot expand)

Day 5

Thus, du must be 0!! 

So,
u
dT u
da
du 

0
T a const dt a T const dt
Day 5

Thus:
u
u
dT

a T const
T a const da

Buuut, Joules experiment also showed that
dT = 0 soooo,
u
0
a T const
Day 5

 This (celebrated) result is known as
Joule’s law, which is strictly true for an Ideal
gas.

Thus,
u
du 
dT
T a const
Day 5/6

 Thus, u = u(T) internal energy is a function
of T only!!!

Q: So what did Joule’s experiment show?

A: Joule’s experiment, separate from Joules’
law showed that heat and mechanical energy
are two forms of the same thing.
Day 6

Next, let’s discuss the concept of specific heat or
specific heat capacity:

Suppose we add small amount of heat (dq)

Temp changes from T to T + DT

Then, if no phase change occurs, the ratio of:

dq/dT = Constant
Day 6

This is the definition of specific heat (which is unique
to each substance or gas).

Add lots o’ heat and small dT

then dq/dT

Example: Metal – heat is dispersed/conducted
readily throughout the substance. (add little heat get
larger dT) small heat capacity, small Cp
= some big number
Day 6

Wood – large heat small dT inside (specific
heat large)

Then a:

Conductor  small specific heat
Insulator  large specific heat

Day 6

We are interested in the behavior of gasses.

The specific heat of a gas:

dq/dT = Constant
Day 6



will have different values depending on what
happens to p or a as gas receives heat, thus:
q
Cv = T a (specific heat at constant Volume) =
717.59 J kg-1 K-1

q
Cp = T p (specific heat at constant pressure) =
1004.63 J kg-1 K-1

(learn ‘em, live ‘em, love ‘em)

Day 6

Important Points:

 Thus, for Cv gas cannot expand as heat is added
(constant Volume), but Pressure can increase

 Thus, for Cp pressure is kept constant as heat is
added, but gas is allowed to expand (a increase)
Day 6



But, as we saw…..
Cp > Cv
for any given gas (why?)
A: since as heat is added and temperature rises, but
pressure is kept constant, then some heat added will
have to be expended to do the work, as material
expands against the constant pressure environment!
Day 6

In plain English: A larger quantity of heat must be
added to the same amount of gas at constant P
(press work term gets some), than if the volume kept
constant (no pressure work term).

***Specific heats are important since for ideal gas
they are constants.
Day 6

If constant volume (press work term =0)

Dq = du

if u = u(T), then,

Cv = Dq/dT = CvdT = Dq (substitute
upstairs).
Day 6

Or….

Du = CvdT

u = CvT + Const. (having applied the snake)

Now the celebrated result! ……
Day 6

The 1st Law……
dT
da

Q  Cv
p
dt
dt

(1st law [pure partitioned form] is in terms of
P, a, and T!!!)

What about const pressure process?
Day 6

Ugh! Let’s see, there’s “p” above, so
maybe:
d
dp
da
dT
 pa  RT   a  p  R
dt
dt
dt
dt

(Q: where’s the other RHS term?)
Day 6

Insert into 1st law and:
dq
dT
dT
dp
 Cv
R
a
dt
dt
dt
dt

 If constant P process,

then,……..

and,

dp = 0
dq = (Cv + R) dT
Day 6/7





or,
dq/DT = Cv + R (which is definition of Cp!)
So,
Cp > Cv as expected (as reasoned out before!).
717.59 + 287.04 = 1004.63.
Day 7

Adiabatic process (Dq/Dt = 0)

Recall “Adiabatic” means that no heat is added or
removed to a system (parcel) (from the surrounding
air).

Thus, “Diabatic” processes describe heat input or
extraction. This heat must be distributed
(apportioned) between internal energy and pressure
work!
Day 7

If a “parcel” is in motion, we call this adiabatic motion. This says
absolutely nothing about how the temperature of the parcel
changes recall these are different concepts.

Important!!! Dq/Dt = 0 does not mean dT/dt = 0 (adiabatic and
isothermal processes are not the same!)

Many atmospheric motions can approximate adiabatic motion
for a few minutes, up to one day, as long as we’re away from
precipitation areas.
Day 7

Parcel theory

In atmosphere: molecular heat transfer important
only for first few centimeters of air closest to the
ground. Vertical mixing is mainly the result of bulk
tranfer by “air parcels” of varying sizes (depending
on the scale we’re talking about).

So, let’s consider a parcel that’s of infinitesimal size!
(Compared to whole atms)
Day 7

Assumptions made:

1) parcel is thermally insulated from it’s environment, so that
temp changes adiabatically as it rises or sinks.

2) The parcel is at exactly the same pressure as it’s
environment (at same level), Pparcel = Penv

3) Environment is in hydrostatic balance

4) Moving slowly enough such that the KE is small.
Day 7

**For real air parcels one or more of these
assumptions are likely to be in violation,
however with this simple conceptual model, it
is easier to understand some of the physical
processes that influence vertical motions and
mixing in the atmosphere.
Day 7

Relate 1st law to air movement in vertical

 If we know that p = p(x,y,z,t) and dp/dt = w

Also, for synoptic scale features V >>> w

such that
dp/dt = -rg dz/dt
Day 7

So for synoptic features in hydrostatic
balance:
ww
 or 
w   rgw
Day 7/8

Assuming this we can write first law in the following
form:
Dq
dT
dz
 cp
g
Dt
dt
dt
 or 
Dq d
 CpT  gz 
Dt dt

 This is another form of the 1st law valid for
synoptic – scale motions where

M = St = CpT + gz = CpT + F
Day 7/8

St is called the “dry static energy” (Internal +
Potential energy).

F  Is geopotential (m2 s-2)

CpT  is enthalpy or sensible heating

This is a form of the first law that relates temperature
change to height change!
Day 8

In q coordinates this is called the
Montgomery Streamfunction (M) (and we
shall talk about this more in 4320/7320)

Stream function defines air motions
assuming a balance condition, thus these are
streamfunctions for adiabatic atmosphere.
“M” is used in PV framework!
Day 8

The dry adiabatic process change rate and
the dry adiabatic lapse rate!

Consider the 1st law written in the following
form:
Dq d
 CpT  gz 
Dt dt

For an adiabatic process:
d
CpT  gz   0
dt
Day 8

Thus for an air parcel dry static energy is conserved!

And, (if we invoke “il serpente”)


CpT + F = Constant
Now, Let us consider a parcel of air undergoing
adiabatic motion, how do we arrive at the dry
adiabatic lapse rate?
Day 8

Since:

F =g dz/dt,

And,

CpdT/dt + gdz/dt = 0,
Day 8

Which means that,

CpdT/dt = -gdz/dt

Thus we get

dT/dz = -g/Cp = Gd

(the dry adiabatic lapse rate! – which is valid in any
planetary atmosphere!)
Day 8

What assumptions did I make in deriving Gd?

hydrostatic balance, and

air is undergoing adiabatic vertical motion.

Thus as a parcel rises, temp changes by a constant
rate g/Cp which is approximately 10 K/km
Day 8

Or: 9.81 / 1004.63 which is 9.76 C(K) / km. (T
decreases/increases as parcel rises/sinks)

In English please?!


5.4 F / 1000 ft (good question for test or HW!)
Recall, for adiabatic process, no heat is gained or
lost, but if we move up 1 km, temp certainly does
change!
Day 8

The notation dT/dZ refers to vertical motion
following a particular parcel of Air!!

***Recall that in Atmospheric Science dT/Dz
is minus, but it’s customary to refer to a drop
in T with height as positive, so 10 C/ km
represents a 10 degree drop with 1 km
height.
Day 8

This compares with the observed Lapse rate of 6 to
7 degrees C / km. Thus parcels moving upward at
the dry adiabatic lapse rate quickly become colder
than their surroundings.

Q: What does this say about the Atmosphere?

A: So if the atmosphere is dry much of the time, then
much of the time, the troposphere is inherently
stable!
Day 8

Now suppose the air parcel contains water vapor:

a. Water vapor will condense as parcel rises (main
cloud producing mechanism).

b. Water vapor condenses releasing latent heat to
the parcel. (Adding heat not same as changing
temperature)
Day 8

1st Law:

So when heat is added to the parcel, some goes into increasing
the temperature, and some into changing the pressure.

Q: So what happens if water vapor begins to condense within
the rising air parcel above?

A: The latent heat of condensation added to parcel,
(counteracting the adiabatic cooling) and the rate of temp
decrease will be something less than 9.76 C/km
dT
dp
Q  c p
a
dt
dt
Day 9

Modifying the 1st Law  Entropy and Potential
Temperature

(we need to show that differentials on RHS are
exact)

Suppose an air parcel undergoes some process or
change, which we can show as a curve form an
initial state A to final state B on a thermodynamic
diagram.
Day 9

**Recall thermodynamic diagram does not show the actual
Physical Path in Time or Space (it’s a point location!)

Thus, the process is reversible! (In higher level classes,
reversible will be referred to as a “Newtonian” one)

Irreverisble process  “Hamiltonian” or sensitively dependent
on intial conditions, “chaotic”.

We want to evaluate the loss or gain of heat (dq – recall no
analytic expression) as parcel goes from A to B on the diagram.
Day 9


We write first law:
Dq
dT
da
 cv
p
Dt
dt
dt
Change in heat during the entire “lifting” or
“sinking” process (finite process), is (invoking
the snake):
B
B
B
B
Dq   Cv
A
dT
da
dT
da
 p
 cv 
 p
dt A dt
dt A dt
A
Day 9

Thus we go from point at (T at this point and P, both
known!) to point B, where we know the P, but NOT T!
We will predict this.

When an integral of a quantity along any curve
depends only on its value at the endpoints of the
B
curve, like:
dT
cv 
dt
A

Recall: that this is the definition of an exact
differential, hence dT/dt is an exact differential!
Day 9

However, pda/dt depends on the history and
“initial conditions” of the parcel, thus it is path
dependent (not an exact differential).

The parcel could go through a phase change
of a gas, etc. which also changes the very
makeup of the parcel. Also, T functionally
depends on P and a). The path from A to B
makes a big difference!
Day 9

These are irreversible processes!

Mathematically speaking, this integral is “improper”. We can’t
evaluate it. (Bummer!  )

Important! ***Thus, the adiabatic problem is really a problem in
prediction! Recall, we said we could use the 1st Law in a
predictive capacity)

So, let’s revisit problem just discussed. It’s not an “esoteric”
problem, but a real problem, i.e. for numerical modeling and
weather forecasting.
Day 9

OK, Suppose we know the state of a parcel at start
(A):

So we know:

At point A  P = Pa and T = Ta

We want to predict what T will be in (for example 12hr), so we want to calculate Tb
for the air parcel.
Day 9


Suppose we can deduce where (B) will be
approximately in 12-hr by some model forecast of Pb.
At point B  P = Pb and T = ????

Thus, our problem is:

Given:
Ta and Pa and Pb
(Forecast) get:
Tb.

Day 9

We have to ass\u\me:

Parcel undergoes an adiabatic and reversible
process from A to B.

Use unmodified 1st law: (to get:)
Day 9

Whoa!
dT
da
Q  Cv
p
dt
dt
B
B
dT
da
Dq   Cv
 p
0
dt A dt
A
dT
da
Cv


p
A dt A dt
B
B
B
Cv (Tb  Ta )    p
A
da
dt
so
1
da
Tb  Ta   p
Cv A dt
B
Day 9/10

**This says there’s a tradeoff between internal
energy and pressure work!

To evaluate Tb we need to know path on diagram
from A to B, we know Pa and have a prediction for
Pb, but what is the path???? We can’t solve the
problem! Double Bummer!! 

Let’s try alternate form Cp form of 1st law!
Day 9/10

Q: will this help?

A: NO! Doesn’t help since (again adiabatic
(dq =0))

So we get:
Day 10

Whoa again!
dT
dp
Q  Cp
a
dt
dt
B
B
dT
dp
Dq   Cp
 a
0
dt A dt
A
B
B
dT
dp
Cp

a
A dt A dt
B
dp
Cp (Tb  Ta )   a
dt
A
so
B
1
dp
Tb  Ta 
a
Cp A dt
Day 10

Still don’t know the integral’s path!!! Arrrgh!!!!

Modification of 1st Law into “perfect” differential
form
dq
dT
da
 cv
p
dt
dt
dt

The 1st Law:

We rewrite the 1st law in terms of a quantity called
“change of specific entropy”
Day 10

We rewrite the 1st law in terms of a quantity
called “change of specific entropy”

Ds/Dt = dq/dt x (1/T)

So we need to deevide 1st Law by T:
ds 1 dq
1 dT P da

 cv

dt T dt
T dt T dt
Day 10

RHS: P/T ??? Why that’s just R/a!!!
ds 1 dq
1 dT R da

 cv

dt T dt
T dt a dt

And:
ds
d
d
 cv ln T   R ln a 
dt
dt
dt
and
ds d
 cv ln T  R ln a 
dt dt
Day 10

Now we have perfect differential form!

Works with other form as well:
ds
d
d
 c p ln T   R ln p 
dt
dt
dt
and
ds d
 c p ln T  R ln p 
dt dt
Day 10

Important!!!

So now you see that dq (change in heat) is
not an exact differential, however, change in
specific entropy is exact!!!

All we did was divide by T, (to eliminate T
dependence in the pressure work term.)
Day 10

How does this help? Unmodified form, we
could not do integral since we did not know
path from A to B.

Modified form of first law to get Tb

Change in specific entropy:
Day 10

Here it is?
ds 1 dq
d
d
d

 Cv ln T  R ln a  cv ln T  R ln a 
dt T dt
dt
dt
dt
Tb
ab
Ds   d cv ln T  R ln a   cv ln  R ln
Ta
aA
A
B

The “improper” integral is gone!! We can
assume adiabatic and continue on!!!
Day 10

Use ideal gas law to get rid of a.
Tb
 TbPa 
)   R ln

Ta
 TaPb 
Tb
Pa 
 Tb
cv ln   R ln  ln 
Ta
Pb 
 Ta
Tb
Tb
Pa
cv ln  R ln   R ln
Ta
Ta
Pb
Tb
Pa
Cp ln   R ln
Ta
Pb
Tb  R Pa
ln 
ln
Ta Cp Pb
then
cv ln(
Pa
Tb  Ta 
 Pb 
R
Cp
Pb
 Ta 
 Pa 
R
Cp
Day 10
Pa
Tb  Ta 
 Pb 
•Hint:
R
Cp
Pb
 Ta 
 Pa 
R
Cp
???
What’s this look like?
•So we get to with our prediction
problem:
•Tb = Ta(Pb/Pa)k
Day 10/11

The resolution of our “problem in prediction”

So the change in specific entropy was calculated
from knowing T and a at the start and end of the
process, but NOT knowing the history of the process,
or the curve itself.

Important!!! In our case: ds = dq = 0, so the process
was reversible!!! (Able to be displayed as a curve on
the diagram). Reversible No phase changes!!!
Day 11

Again:

Reversible (Newtonian processes) path can go both
ways, final state not dependent on specification of
initial conditions. (adiabatic process) potential temp
graphically

Irreversible (Hamiltonian processes), path
dependent, final state highly dependent on the
specification of initial conditions. (Diabatic process)
equivalent potential temp graphically
Day 11

This is the whole crux of numerical weather prediction, and the
whole concept behind ensemble forecasting. (Take intro. to
Chaos theory).

An asside: If you have an adiabatic and irreversible process,
then the entropy can increase OR:

Ds > dq/T This is the second law of thermodynamics.

So the power of rewriting the first law in exact form is that
knowing the initial state, we can get to the final state by
eliminating our dependence on path (Makes the modeler’s job
much simpler).
Day 11

Ok, now suppose we have state A, where we have Ta
and Pa and we end up at state B where Tb = Tref,
and Pb = 1000 hPa, we have

T(1000 hPa) = T (1000 hPa/p)k

**This is our definition of potential temperature! Bring
the parcel of air adiabatically (and reversibly) from
it’s state down to the reference state.
Day 11

***This is the true derivation! Even though you can
take a shortcut, via the first law in inexact form, we
escape the nasty consequences of improper
integrals from the fact that ds = dq = 0.

Adiabatic process = isentropic process (all defined
for a dry atmosphere, however, for an unsaturated
atmosphere the errors are small even if there is H2O
vapor, T is close enough that we don’t have to be
accurate)
Day 11

Now prove that isentropic process means constant potential
temperature:

(In meteorology there is always more than one way to “skin a
cat”)

We want to show dT/DZ = -g/Cp

We did this using the dry static energy relationship last time:
CpT + F = M

Day 11

Now potential temperature equation:

q = T(Po/P)k

Take the natural log:
R
R
ln q  ln T 
ln Po 
ln p
Cp
Cp
Day 11

Then take d/dt of this expression.

Q: What happens to middle term on RHS?

A: It disappears! Why?
1 dq 1 dT
R dp


q dt T dt pCp dt
Day 11

If adiabatic, then what about LHS???

After a bit ‘o algebra:


dT / dp = a /Cp
This is the adiabatic lapse rate in (x,y,p)
coordiates.
Day 11




I can take hydrostatic balance relationship:
dp / dz = -rg
and we can apply the chain rule! (Can you see it?) a
and r are recipricols of each other.
dT/dz = - g /Cp
Day 11

Thus for an isentropic process: theta is conserved!
(Any process following dry adiabat)!

The potential temperture relationship is extremely
useful, and it’s a very powerful way to examine
atmospheric processes (in isentropic coordinates
and assuming adiabatic). Since most processes on
synoptic scale are close to isentropic.
Day 11

The first law and entropy gives us a powerful
compliment to eqn. of state, and hydrostatic approx.
This is the whole foundation of “PV thinking”

In fact Dr. Rainer Bleck (U Miami) suggested at a
meeting that if meteorologists scrap everything
we’ve been taught and just learn isentropic and PV
analysis, we’d know all we need to know about the
atmosphere.
Day 11

To show that the first law and eqn of state are
complimentary, the eqn. Of state can take on
“simplified” forms under the assumption of adiabatic,
and hydrostatically balanced flows:

They are:

TP-k = const.
constant
Pa(Cp/Cv) = constant
aT(Cv/R) =
Day 11

The first coming from Tp-k = qPo-k = constant.

But remember, there’s several ways to get each!

Bonus: Show mathematically that you can convert
mixing ratio to vapor pressure by following a T line to
622 hPa!
Day 12

Bluestein pp 195 – 200 on dry
thermodynamics

Given the first Law in form:
dT Q a
  w
dt c p c p
 or 

T
T a
Q
 Vh   hT  w
 w
t
p c p
cp
Day 12

The equations
dT Q a
  w
dt c p c p
 or 

T
T a
Q
 Vh   hT  w
 w
t
p c p
cp

(1)
(2) (3) (4)
Day 12

where a/Cp is the dry adiabatic lapse rate Gd,
and the partial of T w/r/t to p is, of course the
environmental lapse rate Ge.

Recall: We’ve discussed using this form to
estimate “vertical motion”:
Day 12

 We can assume adiabatic conditions
again, although this expression would also
apply to a non-adiabatic atmosphere as
well):

Thus the local rate of change of temperature
is due to:




T
a T
Q
 Vh   hT    w 
t
cp
 c p p 
Day 12




The Equation:

 a T 

T
Q
 Vh   hT    w 
t
cp
 c p p 
A
B
C
Temperature advection term (horizontal)
Vertical temperature advection (now in term B)

Adiabatic temperature changes due to vertical motion and
atmospheric stability (term B)

Diabatic heating term C.
Day 12

Now let’s get a closer look at: Static stability
(term B):
a T
S

cp

p
and let’s use the relationship for potential
temperature:

Po
q T
P
Day 12

“logrithmic differentiate” (and a bit o’
algebra):
1 q 1 T
R p


q t T t c p p t
becomes
T q T RT



q p p pc p
Day 12

We get a relationship for static stability, which we
name “S”
T q  a T 
S 
     Gd  Ge 
q p  c p p 

This is static stability (e.g, Zwack and Okossi, 1986;
Lupo et al., 1992)! The difference between the dry
adiabatic and environmental lapse rate! Don’t we
calculate this graphically?
Day 12

***Also: static stability defined as “sigma” s

You’ll find this form of the First Law in
Bluestein (1992):

T
P 
Q

 Vh   hT   s w 
t
cp
R 
where
s 
R T q
p q p
Day 12



If the atmosphere is “dry-neutral” then
obviously:
S = 0  Since Ge = Gd  a/Cp = dT/dp
Let’s take a look at Static stability (S) vs.
Vertical motion (w)
Day 12

To do this, let’s isolate these two variables, so
assume that:

The local rate of change in T is constant (C).
The temperature advection is zero.
The diabatic heating is zero.




Q: Is this unreasonable?
A: maybe, maybe not.
Day 12



Then the First Law becomes:
C = Sw 
w = C/S
Important Definition: Static Stability “S” can
be defined as the “resistivity” of the
atmosphere to vertical overturning (motion)!
Day 12

 How to interpret: for the same amount of heating,
a larger (smaller) stability (more[less] stable air)
resists vertical motion and produces a smaller
(larger) w.

So, in the case of large static stability, expansion and
compression of air overwhelm, vertical temperature
advection, or the vertical advection may act in the
same sense as T (cool air under warm)!
Day 12

Q: Where in the atmosphere (Homosphere)
is S very large and consequently, vertical
motions are very small?

We have looked at the 1st Law in (x,y,p)
coordinates.
Day 12

In (x,y, q) coords. It’s:

q
q q dq
 Vh   hq  w

t
p c pT dt
A
B
C

Where term;

A - is the advection of potential Temp.
B - is the vertical advection of potential temp
(stability term)
C - Is the diabatic heating term



Day 12

This version is valid for synoptic scale
process only. On smaller scales, on scale of
convection, where hydrostatic balance does
not hold, we must use (x,y,z) thermo dynamic
equation.
Day 13

Baroclinic atmosphere vs. Barotropic
atmosphere.

An atmosphere in which density is a function
of pressure and temperature is called a
baroclinic atmosphere. P (mass) and r
(thermal) fields cross to form solenoids:
Day 13

…soleniods (avoid the ‘noid?)

(Gradient of density and pressure not
parallel).
Day 13

This is where all the work is done in the
atmosphere, and this is where vorticity is
generated!

 This is where Available potential energy is
converted to Kinetic by cyclones and
anticyclones! (Midlatitudes)
Day 13

Some math…
r  r  p, T 
 or 
0  p  r
 or 

V
0
p
Day 13

An atmosphere in which density is a function
of pressure only is a barotropic atmosphere.
r  r  p

A Barotropic atmosphere is isothermal, thus
there is NO advection of temperature. There
are no vertical wind shears and NO
solenoids (avoid the ‘noid!).
Day 13

(Gradient of density and pressure parallel)
(Tropics may nearly mimic at times)
Day 13

Absolute vorticity conserved! No Available
Potential Energy. A barotropic atmosphere is
a steady state, basic state atmosphere.

Equivalent Barotropic  isotherms are
parallel to the pressure lines. There are
horizontal temperature advections.
Day 13

 However there is vertical shear. (thus density and
pressure gradients nearly, but not quite parallel.

Some examples of this type of atmosphere:


The Tropics
cutoff lows
Blocking anticyclones

We’ll talk more about these ideas in dynamics.

Day 13

The thermodynamics of Moist air

(Read Bluestein 200 – 223, and Hess ch 4 and 5)

The equation of state for an atmosphere of water
vapor only (A “water world?” – sorry Kevin Costner):

Pv = R*/mv rvTv
Day 13

Let’s call vapor pressure Pv 

and Rv is 

R*/mv = 8314.3 J/K kg / 18.016 = 461.5 J/K
kg
e
Day 13

Thus, the equation of state is: e av=RvT or
e = rvRvT

 This is the equation of state for water
vapor itself or as a constituent of moist air.

Next, consider moist air + dry air, and now
the parcel is saturated or e= es
Day 13

Then this vapor equation is:

es av=RvT or es = rvRvT

Saturation or Equilibrium Vapor Pressure (es)

 “es” is a function of temperature only and not
dependent on the pressure of the other gasses
present
Day 13

The concept of equilibrium vapor pressure
over a plane of pure water (does the
atmosphere “hold” water vapor?):
Day 13

The Variation of es (es over water and es over ice
– or “on the rocks”) with temperature:
Temperature
esw(hPA)
esi (hPa)
esw - esi
-20 C
1.25
1.03
0.22
-10 C
2.86
2.60
0.26
0C
6.11
6.11
0
10 C
12.27
n/a
20 C
23.37
n/a
30 C
42.45
n/a
40 C
73.77
n/a
Day 13

Graph here:
Day 13
Day 13