lecture 6 adiabatic processes. definition process is adiabatic if there is no exchange of heat...
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Lecture 6Lecture 6
Adiabatic ProcessesAdiabatic Processes
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DefinitionDefinition
Process is Process is adiabaticadiabatic if there is no if there is no exchange of heat between system and exchange of heat between system and environment, i.e.,environment, i.e.,
dq = 0dq = 0
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Work and Temperature (General)Work and Temperature (General)
First law: dU = dQ – dWFirst law: dU = dQ – dW
Adiabatic process: dU = -dWAdiabatic process: dU = -dW
If system If system does workdoes work (dW > 0), dU < 0 (dW > 0), dU < 0 system coolssystem cools
If work is done If work is done on on system (dW < 0), dU > 0system (dW < 0), dU > 0
system warmssystem warms
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Work and Temperature (Ideal Gas)Work and Temperature (Ideal Gas)
Adiabatic process: cAdiabatic process: cvvdT = - pddT = - pd
Expansion (dExpansion (d > 0) > 0) dT < 0 (cooling) dT < 0 (cooling)
Contraction (dContraction (d < 0) < 0) dT > 0 (warming) dT > 0 (warming)
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Relationship between T and pRelationship between T and p
dpdqdTc p
dq = 0
dpdTc p But,
p
RT
[Eq. (4) from last lecture.]
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ThusThus
dpp
RTdTc p
p
dp
c
R
T
dT
p
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Adiabatic TransitionAdiabatic Transition
Suppose system starts in state with Suppose system starts in state with thermodynamic “coordinates” (Tthermodynamic “coordinates” (T00, p, p00))
System makes an adiabatic transition to System makes an adiabatic transition to state with coordinates (Tstate with coordinates (T11, p, p11))
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IntegrateIntegrate
1
0
1
0
T
T
p
pp p
dp
c
R
T
dT
11 lnlnp
pp
T
T oop
c
RT (1)
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ContinueContinue
(1) 0101 lnlnlnln ppc
RTT
p
0
1
0
1 lnlnp
p
c
R
T
T
p
(2)
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ReviewReview
xaxa lnln
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Thus,Thus,
(2) becomes
(3)
pc
R
p
p
T
T
0
1
0
1 lnln
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Thus,Thus,
(3) becomes(3) becomes
pc
R
p
p
T
T
0
1
0
1
Poisson’s Equation
(4)
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Dry AirDry Air
2859.01004
0.28711
11
KkgJ
KkgJ
c
R
p
2859.0
0
1
0
1
p
p
T
T
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ExerciseExercise
pp00 = 989 hPa = 989 hPa
TT00 = 276 K = 276 K
pp11 = 742 hPa = 742 hPa
TT11 = ? = ?
Answer: TAnswer: T11 = 254 K = 254 K
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ExerciseExercise
pp00 = 503 hPa = 503 hPa
TT00 = 230 K = 230 K
pp11 = 1000 hPa = 1000 hPa
TT11 = ? = ?
Answer: 280 KAnswer: 280 K
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Potential TemperaturePotential Temperature
Let pLet p00 = 1000 hPa = 1000 hPa
Remove the Remove the subscripts from psubscripts from p11
and Tand T11
Denote TDenote T00 by by pcR
p
hPaT
/1000
is called the potential temperature
pcR
p
PT
/
0
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Physical MeaningPhysical MeaningInitial state: (T, p)Initial state: (T, p)Suppose system makes an adiabatic transition Suppose system makes an adiabatic transition to pressure of 1000 hPa to pressure of 1000 hPa New temperature = New temperature =
Potential temperature is the temperature a Potential temperature is the temperature a parcel would have if it were to expand or parcel would have if it were to expand or compress adiabatically from its present compress adiabatically from its present pressure and temperature to a reference pressure and temperature to a reference pressure level. Po = 1000 mb.pressure level. Po = 1000 mb.
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Physical MeaningPhysical MeaningRemoves adiabatic temperature changes experienced during vertical motion
ºC and K are interchangeable; best to convert it to K when making calculations such as differences.
is invariant along an adiabatic path
adiabatic behavior of individual air parcels is a good approximation for many atmospheric applications…from small parcels to larger convection.
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AdiabatsAdiabats
Let Let be given be given
Re-write last equation:Re-write last equation:
pc
R
hPa
pT
1000
In the T-p plane, this describes a curve.
Curve is called a dry adiabat.
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Dry AdiabatsDry Adiabats
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= 290 K= 290 K
Initial state:
T = 290.0 K, p = 1000 hPa
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Reduce pressure to 900 hPaReduce pressure to 900 hPa
New temperature:
T 281 K
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ExerciseExercise
Calculate T to nearest tenth of a degreeCalculate T to nearest tenth of a degree
2859.0
1000
hPa
pT
K
K
4.281
1000
900290
2859.0
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Reduce pressure to 700 hPaReduce pressure to 700 hPa
New temperature:
T 262 K
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ExerciseExercise
Calculate new T to nearest tenth of a Calculate new T to nearest tenth of a degreedegree
Answer: 261.9 KAnswer: 261.9 K
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Adiabatic ProcessesAdiabatic Processes
In the T-p plane, an adiabatic process can In the T-p plane, an adiabatic process can be thought of as a point moving along an be thought of as a point moving along an adiabat.adiabat.
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The Parcel ModelThe Parcel Model
1.1. An air parcel is a hypothetical volume of An air parcel is a hypothetical volume of air that does not mix with its surroundingsair that does not mix with its surroundings
1.1. Parcel is a Parcel is a closed closed system.system.
2.2. Parcel moves adiabatically if there is no Parcel moves adiabatically if there is no exchange of heat with surroundings.exchange of heat with surroundings.
1 and 2 1 and 2 parcel is parcel is isolatedisolated Parcel doesn’t interact with surroundings.Parcel doesn’t interact with surroundings.
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Rising & Sinking ParcelsRising & Sinking Parcels
If a parcel rises adiabatically, its pressure If a parcel rises adiabatically, its pressure decreasesdecreases parcel coolsparcel cools
If a parcel sinks adiabatically, its pressure If a parcel sinks adiabatically, its pressure increasesincreases parcel warmsparcel warms
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Movie: “The Day After Tomorrow”Movie: “The Day After Tomorrow”
Premise: global warming produces Premise: global warming produces gigantic stormsgigantic storms
In these storms, cold air from upper In these storms, cold air from upper troposphere is brought down to surface, troposphere is brought down to surface, causing sudden coolingcausing sudden cooling (People freeze in their tracks!)(People freeze in their tracks!)
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Problem With PremiseProblem With Premise
As air sinks, it WARMS!As air sinks, it WARMS!
Suppose air at height of 10 km sinks Suppose air at height of 10 km sinks rapidly to surfacerapidly to surface
Pressure at 10 km Pressure at 10 km 260 hPa 260 hPa
Temperature Temperature 220 K 220 K
If surface pressure If surface pressure 1000 hPa, what is 1000 hPa, what is air temperature upon reaching surface?air temperature upon reaching surface?
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SolutionSolution
286.0
260
1000220
hPa
hPaKTsfc
K323C50F122
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Height Dependence of THeight Dependence of T
pcR
p
pT
/
0
*
Start with
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Parcel TemperatureParcel Temperature
Consider a parcel with pressure Consider a parcel with pressure pp and and temperature temperature TT..
Assume the parcel rises adiabaticallyAssume the parcel rises adiabatically is constantis constant
Goal: Determine dT/dz.Goal: Determine dT/dz.
Method: logarithmic differentiationMethod: logarithmic differentiation
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Step 1Step 1
Take log of both sides of *Take log of both sides of *
0
/
0
lnlnln
lnlnln
pc
Rp
c
R
p
pT
pp
cR p
Constants
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Step 2: DifferentiateStep 2: Differentiate
pdz
d
c
RT
dz
d
p
lnln
dz
dp
pc
R
dz
dT
T p
1 **
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Step 3: Hydrostatic EquationStep 3: Hydrostatic Equation
gdz
dp
Substitute into **
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Step 4Step 4
gpc
R
dz
dT
T p
1
pc
g
p
RT
dz
dT
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ExerciseExercise
Simplify the expression for dT/dz.Simplify the expression for dT/dz.
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ResultResult
pc
g
dz
dT
Constant!
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z
Parcel temperature
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ConclusionConclusion
A parcel that rises adiabatically cools at A parcel that rises adiabatically cools at the rate g/cthe rate g/cpp
A parcel that sinks adiabatically warms at A parcel that sinks adiabatically warms at the rate g/cthe rate g/cpp
Exercise: Calculate g/cExercise: Calculate g/cpp(in K(in Kkmkm-1-1)) Answer: 9.8 KAnswer: 9.8 Kkmkm-1-1
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Dry-Adiabatic “Lapse Rate”Dry-Adiabatic “Lapse Rate”
p
d
pd c
R
c
R
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Important NoteImportant Note
Previous discussion assumed that there Previous discussion assumed that there was no condensation.was no condensation.
We will look at the effect of water vapor We will look at the effect of water vapor next.next.