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Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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Page 1: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Atmospheric Science 4310 / 7310

Atmospheric Thermodynamics - IIByAnthony R. Lupo

Page 2: 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:

Page 3: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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

So only forces parallel to the distance travelled do work!

dsFW

Page 4: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 1

Then,

dtds

FdtdW

or

dsFdW

Page 5: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 1

But, we know that:

Pressure = Force / Unit Area

So then,

Force = P x Area

Page 6: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 1

Total work increment now:

Well,

Area x length = Volume

soo………………… A * ds = dVol

dsAreaPdW

Page 7: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 1

Then we get the result:

Work :

Let’s “work” with Work per unit mass:

Thus, we can start out with volume of only one 1 kg of gas!!!

PdVdW

mdV

PmdW

Page 8: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 1

And we know that:

= Vol/m

then……..

Note the “heavy D”

PdDW

Page 9: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 1

So….

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 2 > 1

2

1

PdW

or

PdDW

Page 10: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 1

Parcel does work expanding against environment!

But if the parcel undergoes contraction (2 < 1)

Environment is doing work on the parcel!

Page 11: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 1/2

A derivation more relevant to an air parcel

Given a spherical parcel of air with radius R, Surface Area = 4r2 and the volume is 4/3r3.

Page 12: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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

Page 13: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 2

Well,

Dist = dr

And;

Area = A + dA

Page 14: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 2

As with the “expanding piston”

DW = p*A*dr = p * dV

Recall from Calc II (or is it physics?)

dV = 4 r2 dr

Page 15: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 2

Then we’ll deevide by the unit mass again to get:

dw = p d

Now the short comings in this problem are obvious:

1. (Surf)Area = A + dA = 4 r2 + 8 r dr

Page 16: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 2

2. The parcel expands and p decreases;

so: p = p + dp

So….. the REAL problem is:

dW = (p+dp)(A+dA) dr

Page 17: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 2

Which “foils” out to,

DW = pAdr + dpAdr + pdAdr + dpDAdr

then assume;

Dw = p d + dp d + p dA dr + dp dA dr

Page 18: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 2

Which then reduces to (by scale analysis):

Dw = p d

Page 19: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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, (), or pressure for a parcel.

Page 20: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 21: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)

VdVmassdtdKE

or

velocitymassKE

221

21 2

Page 22: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 2

PE (potential energy) is the energy of position relevant to some reference level, or within some potential (e.g., gravitational) field.

dtd

dtdz

g

wheredtdz

gmassdtdPE

or

heightgmassPE

*

Page 23: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)!!!

Page 24: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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!):

Page 25: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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”

heat = KE + PE + IE + Work + E&M

Page 26: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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).

Page 27: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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

KE = ½ (50 m/s)2 – ½ (10 m/s)2 or 1.2 x 103 m2 sec -2

Page 28: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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:

IE = Cv T = 717 J K-1 kg-1 x 20 K

or 1.4 x 104 m2 sec-2

Page 29: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 3

Then, let’s consider the change in PE at, oh, say 500 hPa,

So,

PE = g*z = 10 m s-2 * 60 m = 600 m2 s-2

Page 30: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 3

Pressure work;

We have dw = p d and p = RT so dW = (RT/) d = RT d[ln()]

Since T = Const, we can write:

= ln 0.1 * 287.04 J/kg K * 280 K

W = 8.47 x 103 m2 s-2

Page 31: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)?

Page 32: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 3

Well,

KE = KE(horiz) + KE (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)

Page 33: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 3

The Horizontal pressure gradient is responsible for changes in wind speed.

The pressure gradient is in KE and Wh (which is horizontal wind times distance covered)

Then, KE must cancel with Wh!

How about Wv?

Page 34: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 3

Well, PE = gdz

and,

Wv = vert. PGF x distance = - (1/) (dp/dz) (dz)

If hydrostatic balance is assumed to hold then g = -(1/) (dp/dz).

Page 35: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 3

Thus, PE + 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”)

Page 36: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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

Page 37: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)

Page 38: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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

Page 39: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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, , or P?) so we can obtain useful expressions for atmospheric calculations?

Page 40: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.,

Page 41: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 5

Thus, dw = p d.

So the 1st Law is now:

pdduQ

Page 42: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 5

Internal Energy

With the pressure work term expressed in terms of state variables p and , we must now consider that internal energy may be dependent on (T and ) or (T and p) (since p and can be related)

So: Consider that u = u(T,) or u = u(T,p)

Page 43: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 5

The expression

dtdu

dtdT

Tu

duconstTconst

Page 44: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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).

Page 45: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 5

Draw:

Page 46: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)

Page 47: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 5

Thus, du must be 0!!

So,

0

dtdu

dtdT

Tu

duconstTconst

Page 48: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 5

Thus:

Buuut, Joules experiment also showed that dT = 0 soooo,

ddT

Tuu

constconstT

0

constT

u

Page 49: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 5

This (celebrated) result is known as Joule’s law, which is strictly true for an Ideal gas.

Thus,

dTTu

duconst

Page 50: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 51: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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 + T

Then, if no phase change occurs, the ratio of:

dq/dT = Constant

Page 52: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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 = some big number

Example: Metal – heat is dispersed/conducted readily throughout the substance. (add little heat get larger dT) small heat capacity, small Cp

Page 53: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 6

Wood – large heat small dT inside (specific heat large)

Then a:

Conductor small specific heat Insulator large specific heat

Page 54: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 6

We are interested in the behavior of gasses.

The specific heat of a gas:

dq/dT = Constant

Page 55: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 6

will have different values depending on what happens to p or as gas receives heat, thus:

Cv = (specific heat at constant Volume) = 717.59 J kg-1 K-1

Cp = (specific heat at constant pressure) = 1004.63 J kg-1 K-1

(learn ‘em, live ‘em, love ‘em)

Tq

pTq

Page 56: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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 ( increase)

Page 57: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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!

Page 58: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 59: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 6

If constant volume (press work term =0)

Dq = du

if u = u(T), then,

Cv = Dq/dT = CvdT = Dq (substitute upstairs).

Page 60: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 6

Or….

Du = CvdT

u = CvT + Const. (having applied the snake)

Now the celebrated result! ……

Page 61: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 6

The 1st Law……

(1st law [pure partitioned form] is in terms of P, , and T!!!)

What about const pressure process?

dtd

pdtdT

CvQ

Page 62: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 6

Ugh! Let’s see, there’s “p” above, so maybe:

(Q: where’s the other RHS term?)

dtdT

Rdtd

pdtdp

RTpdtd

Page 63: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 6

Insert into 1st law and:

If constant P process,

then,…….. dp = 0

and,

dq = (Cv + R) dT

dtdp

dtdT

RdtdT

Cvdtdq

Page 64: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 65: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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!

Page 66: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 67: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)

Page 68: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 69: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 70: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 7

Relate 1st law to air movement in vertical

If we know that p = p(x,y,z,t) and dp/dt =

Also, for synoptic scale features V >>>

such that dp/dt = -g dz/dt

Page 71: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 7

So for synoptic features in hydrostatic balance:

gw

or

w

Page 72: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 7/8

Assuming this we can write first law in the following form:

This is another form of the 1st law valid for synoptic – scale motions where

M = St = CpT + gz = CpT +

gzCpTdtd

DtDq

ordtdz

gdtdT

cDtDq

p

Page 73: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 7/8

St is called the “dry static energy” (Internal + Potential energy).

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!

Page 74: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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!

Page 75: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 8

The dry adiabatic process change rate and the dry adiabatic lapse rate!

Consider the 1st law written in the following form:

For an adiabatic process:

gzCpTdtd

DtDq

0 gzCpTdtd

Page 76: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 8

Thus for an air parcel dry static energy is conserved!

And, (if we invoke “il serpente”)

CpT + = Constant

Now, Let us consider a parcel of air undergoing adiabatic motion, how do we arrive at the dry adiabatic lapse rate?

Page 77: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 8

Since:

=g dz/dt,

And,

CpdT/dt + gdz/dt = 0,

Page 78: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 8

Which means that,

CpdT/dt = -gdz/dt

Thus we get

dT/dz = -g/Cp = d

(the dry adiabatic lapse rate! – which is valid in any planetary atmosphere!)

Page 79: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 8

What assumptions did I make in deriving d?

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

Page 80: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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!

Page 81: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 82: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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!

Page 83: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)

Page 84: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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

dtdp

dtdT

cQ p

Page 85: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 86: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 87: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 9

We write first law:

Change in heat during the entire “lifting” or “sinking” process (finite process), is (invoking the snake):

dtd

pdtdT

cDtDq

v

B

A

B

A

B

A

v

B

A dtd

pdtdT

cdtd

pdtdT

Cvq

Page 88: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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 curve, like:

Recall: that this is the definition of an exact differential, hence dT/dt is an exact differential!

B

A

v dtdT

c

Page 89: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 9

However, pd/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 ). The path from A to B makes a big difference!

Page 90: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 91: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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 for the air parcel.

We want to predict what T will be in (for example 12-hr), so we want to calculate Tb

Page 92: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 93: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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:)

Page 94: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 9

Whoa!

B

A

B

A

B

A

B

A

B

A

B

A

dtd

pCv

TaTb

so

dtd

pTaTbCv

dtd

pdtdT

Cv

dtd

pdtdT

Cvq

dtd

pdtdT

CvQ

1

)(

0

Page 95: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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!

Page 96: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 9/10

Q: will this help?

A: NO! Doesn’t help since (again adiabatic (dq =0))

So we get:

Page 97: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 10

Whoa again!

B

A

B

A

B

A

B

A

B

A

B

A

dtdp

CpTaTb

so

dtdp

TaTbCp

dtdp

dtdT

Cp

dtdp

dtdT

Cpq

dtdp

dtdT

CpQ

1

)(

0

Page 98: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 10

Still don’t know the integral’s path!!! Arrrgh!!!!

Modification of 1st Law into “perfect” differential form

The 1st Law:

We rewrite the 1st law in terms of a quantity called “change of specific entropy”

dtd

pdtdT

cdtdq

v

Page 99: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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:

dtd

TP

dtdT

Tc

dtdq

Tdtds

v

11

Page 100: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 10

RHS: P/T ??? Why that’s just R/!!!

And:

dtdR

dtdT

Tc

dtdq

Tdtds

v

11

lnln

lnln

RTcdtd

dtds

anddtd

RTdtd

cdtds

v

v

Page 101: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 10

Now we have perfect differential form!

Works with other form as well:

pRTcdtd

dtds

and

pdtd

RTdtd

cdtds

p

p

lnln

lnln

Page 102: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.)

Page 103: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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:

Page 104: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 10

Here it is?

The “improper” integral is gone!! We can assume adiabatic and continue on!!!

lnlnlnln1

RTcvdtd

dtd

RTdtd

Cvdtdq

Tdtds

B

A Ab

RTaTb

cvRTcvds lnlnlnln

Page 105: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 10

Use ideal gas law to get rid of

Cp

R

Cp

R

PaPb

TaPbPa

TaTb

then

PbPa

CpR

TaTb

PbPa

RTaTb

Cp

PbPa

RTaTb

RTaTb

cv

PbPa

TaTb

RTaTb

cv

TaPbTbPa

RTaTb

cv

lnln

lnln

lnlnln

lnlnln

ln)ln(

Page 106: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 10

Cp

R

Cp

R

PaPb

TaPbPa

TaTb

???

•Hint: What’s this look like?

•So we get to with our prediction problem:

•Tb = Ta(Pb/Pa)k

Page 107: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 10/11

The resolution of our “problem in prediction”

So the change in specific entropy was calculated from knowing T and 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!!!

Page 108: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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

Page 109: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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).

Page 110: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 111: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)

Page 112: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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 + = M

Page 113: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 11

Now potential temperature equation:

= T(Po/P)k

Take the natural log:

pCpR

PoCpR

T lnlnlnln

Page 114: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 11

Then take d/dt of this expression.

Q: What happens to middle term on RHS?

A: It disappears! Why?

dtdp

pCpR

dtdT

Tdtd 11

Page 115: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 11

If adiabatic, then what about LHS???

After a bit ‘o algebra:

dT / dp = /Cp

This is the adiabatic lapse rate in (x,y,p) coordiates.

Page 116: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 11

I can take hydrostatic balance relationship:

dp / dz = -g

and we can apply the chain rule! (Can you see it?) and are recipricols of each other.

dT/dz = - g /Cp

Page 117: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 118: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 119: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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. P(Cp/Cv) = constant T(Cv/R) = constant

Page 120: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 11

The first coming from Tp-k = Po-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!

Page 121: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

Bluestein pp 195 – 200 on dry thermodynamics

Given the first Law in form:

pphh

pp

cQ

cpT

TVtT

or

ccQ

dtdT

Page 122: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

The equations

(1) (2) (3) (4)

pphh

pp

cQ

cpT

TVtT

or

ccQ

dtdT

Page 123: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

where /Cp is the dry adiabatic lapse rate d, and the partial of T w/r/t to p is, of course the environmental lapse rate e.

Recall: We’ve discussed using this form to estimate “vertical motion”:

Page 124: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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:

pphh c

QpT

cTV

tT

Page 125: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

The Equation:

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.

pphh c

QpT

cTV

tT

Page 126: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

Now let’s get a closer look at: Static stability (term B):

and let’s use the relationship for potential temperature:

pT

cS

p

PPo

T

Page 127: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

“logrithmic differentiate” (and a bit o’ algebra):

p

p

pcRT

pT

pT

becomes

tp

pcR

tT

Tt

11

Page 128: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

We get a relationship for static stability, which we name “S”

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?

edp p

Tcp

TS

Page 129: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

***Also: static stability defined as “sigma”

You’ll find this form of the First Law in Bluestein (1992):

pT

pR

where

cQ

RP

TVtT

phh

Page 130: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

If the atmosphere is “dry-neutral” then obviously:

S = 0 Since e = d /Cp = dT/dp

Let’s take a look at Static stability (S) vs. Vertical motion ()

Page 131: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 132: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

Then the First Law becomes:

C = S = C/S

Important Definition: Static Stability “S” can be defined as the “resistivity” of the atmosphere to vertical overturning (motion)!

Page 133: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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) .

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)!

Page 134: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 135: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 12

In (x,y, ) coords. It’s:

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

dtd

TcpV

t phh

Page 136: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 137: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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 (thermal) fields cross to form solenoids:

Page 138: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

…soleniods (avoid the ‘noid?)

(Gradient of density and pressure not parallel).

Page 139: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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)

Page 140: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

Some math…

0

0

,

pV

or

p

or

Tp

Page 141: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

An atmosphere in which density is a function of pressure only is a barotropic atmosphere.

A Barotropic atmosphere is isothermal, thus there is NO advection of temperature. There are no vertical wind shears and NO solenoids (avoid the ‘noid!).

p

Page 142: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

(Gradient of density and pressure parallel) (Tropics may nearly mimic at times)

Page 143: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 144: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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.

Page 145: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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 vTv

Page 146: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

Let’s call vapor pressure Pv e

and Rv is

R*/mv = 8314.3 J/K kg / 18.016 = 461.5 J/K kg

Page 147: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

Thus, the equation of state is: e v=RvT or e = vRvT

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

Page 148: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

Then this vapor equation is: es v=RvT or es = vRvT

Saturation or Equilibrium Vapor Pressure (es)

“es” is a function of temperature only and not dependent on the pressure of the other gasses present

Page 149: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

The concept of equilibrium vapor pressure over a plane of pure water (does the atmosphere “hold” water vapor?):

Page 150: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

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

0 C 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

Page 151: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

Graph here:

Page 152: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13

Page 153: Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

Day 13