atmospheric science 4310 / 7310
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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: - PowerPoint PPT PresentationTRANSCRIPT
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Atmospheric Science 4310 / 7310
Atmospheric Thermodynamics - IIByAnthony R. Lupo
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Day 1
The work done by an expanding gas
Let’s draw a piston:
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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
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Day 1
Then,
dtds
FdtdW
or
dsFdW
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Day 1
But, we know that:
Pressure = Force / Unit Area
So then,
Force = P x Area
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Day 1
Total work increment now:
Well,
Area x length = Volume
soo………………… A * ds = dVol
dsAreaPdW
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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
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Day 1
And we know that:
= Vol/m
then……..
Note the “heavy D”
PdDW
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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
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Day 1
Parcel does work expanding against environment!
But if the parcel undergoes contraction (2 < 1)
Environment is doing work on the parcel!
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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.
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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
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Day 2
Well,
Dist = dr
And;
Area = A + dA
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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
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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
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Day 2
2. The parcel expands and p decreases;
so: p = p + dp
So….. the REAL problem is:
dW = (p+dp)(A+dA) dr
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Day 2
Which “foils” out to,
DW = pAdr + dpAdr + pdAdr + dpDAdr
then assume;
Dw = p d + dp d + p dA dr + dp dA dr
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Day 2
Which then reduces to (by scale analysis):
Dw = p d
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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.
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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.
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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
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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
*
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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)!!!
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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!):
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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
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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).
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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
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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
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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
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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
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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)?
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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)
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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?
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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).
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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”)
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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
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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)
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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
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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?
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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.,
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Day 5
Thus, dw = p d.
So the 1st Law is now:
pdduQ
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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)
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Day 5
The expression
dtdu
dtdT
Tu
duconstTconst
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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).
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Day 5
Draw:
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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)
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Day 5
Thus, du must be 0!!
So,
0
dtdu
dtdT
Tu
duconstTconst
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Day 5
Thus:
Buuut, Joules experiment also showed that dT = 0 soooo,
ddT
Tuu
constconstT
0
constT
u
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Day 5
This (celebrated) result is known as Joule’s law, which is strictly true for an Ideal gas.
Thus,
dTTu
duconst
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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.
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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
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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
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Day 6
Wood – large heat small dT inside (specific heat large)
Then a:
Conductor small specific heat Insulator large specific heat
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Day 6
We are interested in the behavior of gasses.
The specific heat of a gas:
dq/dT = Constant
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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
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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)
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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!
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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.
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Day 6
If constant volume (press work term =0)
Dq = du
if u = u(T), then,
Cv = Dq/dT = CvdT = Dq (substitute upstairs).
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Day 6
Or….
Du = CvdT
u = CvT + Const. (having applied the snake)
Now the celebrated result! ……
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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
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Day 6
Ugh! Let’s see, there’s “p” above, so maybe:
(Q: where’s the other RHS term?)
dtdT
Rdtd
pdtdp
RTpdtd
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Day 6
Insert into 1st law and:
If constant P process,
then,…….. dp = 0
and,
dq = (Cv + R) dT
dtdp
dtdT
RdtdT
Cvdtdq
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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.
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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!
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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.
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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)
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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.
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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.
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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
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Day 7
So for synoptic features in hydrostatic balance:
gw
or
w
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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
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/74.jpg)
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!
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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
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/77.jpg)
Day 8
Since:
=g dz/dt,
And,
CpdT/dt + gdz/dt = 0,
![Page 78: Atmospheric Science 4310 / 7310](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/78.jpg)
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!)
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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
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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!
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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.
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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!
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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)
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/85.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/86.jpg)
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.
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/88.jpg)
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
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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!
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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.
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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
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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.
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/94.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/95.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/96.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/97.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/98.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/99.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/100.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/101.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/102.jpg)
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.)
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/104.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/105.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/106.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/107.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/108.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/109.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/110.jpg)
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.
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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)
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/113.jpg)
Day 11
Now potential temperature equation:
= T(Po/P)k
Take the natural log:
pCpR
PoCpR
T lnlnlnln
![Page 114: Atmospheric Science 4310 / 7310](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/114.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/115.jpg)
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.
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/117.jpg)
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.
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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.
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/120.jpg)
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!
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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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/122.jpg)
Day 12
The equations
(1) (2) (3) (4)
pphh
pp
cQ
cpT
TVtT
or
ccQ
dtdT
![Page 123: Atmospheric Science 4310 / 7310](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/123.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/124.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/125.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/126.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/127.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/128.jpg)
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](https://reader036.vdocuments.net/reader036/viewer/2022062305/56814737550346895db476ca/html5/thumbnails/129.jpg)
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
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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 ()
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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.
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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)!
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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)!
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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.
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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
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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.
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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:
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Day 13
…soleniods (avoid the ‘noid?)
(Gradient of density and pressure not parallel).
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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)
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Day 13
Some math…
0
0
,
pV
or
p
or
Tp
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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
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(Gradient of density and pressure parallel) (Tropics may nearly mimic at times)
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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.
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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.
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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
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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
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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
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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
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Day 13
The concept of equilibrium vapor pressure over a plane of pure water (does the atmosphere “hold” water vapor?):
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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
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Day 13
Graph here:
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Day 13
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Day 13