Lecture 6Lecture 6Adiabatic ProcessesAdiabatic Processes
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
Work and Temperature (General)Work and Temperature (General)
First law: dU = dQ – dWFirst law: dU = dQ – dWAdiabatic process: dU = -dWAdiabatic process: dU = -dWIf 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
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)
Relationship between T and pRelationship between T and p
dpdqdTc p
dq = 0
dpdTc p But,
pRT
[Eq. (4) from last lecture.]
ThusThus
dppRTdTc p
pdp
cR
TdT
p
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))
IntegrateIntegrate
1
0
1
0
T
T
p
pp pdp
cR
TdT
11 lnln p
pp
T
T oop
cRT (1)
ContinueContinue
(1) 0101 lnlnlnln ppcRTTp
0
1
0
1 lnlnpp
cR
TT
p
(2)
ReviewReview
xaxa lnln
Thus,Thus,
(2) becomes
(3)
pcR
pp
TT
0
1
0
1 lnln
Thus,Thus,
(3) becomes(3) becomes
pcR
pp
TT
0
1
0
1
Poisson’s Equation
(4)
Dry AirDry Air
2859.01004
0.28711
11
KkgJKkgJ
cR
p
2859.0
0
1
0
1
pp
TT
ExerciseExercise
pp00 = 989 hPa = 989 hPa
TT00 = 276 K = 276 K
pp11 = 742 hPa = 742 hPa
TT11 = ? = ?
Answer: TAnswer: T11 = 254 K = 254 K
ExerciseExercise
pp00 = 503 hPa = 503 hPa
TT00 = 230 K = 230 K
pp11 = 1000 hPa = 1000 hPa
TT11 = ? = ?
Answer: 280 KAnswer: 280 K
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
phPaT
/1000
is called the potential temperature
pcR
pPT
/
0
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.
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.
AdiabatsAdiabats
Let Let be given be givenRe-write last equation:Re-write last equation:
pcR
hPapT
1000
In the T-p plane, this describes a curve.
Curve is called a dry adiabat.
Dry AdiabatsDry Adiabats
= 290 K= 290 K
Initial state:
T = 290.0 K, p = 1000 hPa
Reduce pressure to 900 hPaReduce pressure to 900 hPa
New temperature:
T 281 K
ExerciseExercise
Calculate T to nearest tenth of a degreeCalculate T to nearest tenth of a degree2859.0
1000
hPapT
K
K
4.2811000900290
2859.0
Reduce pressure to 700 hPaReduce pressure to 700 hPa
New temperature:
T 262 K
ExerciseExercise
Calculate new T to nearest tenth of a Calculate new T to nearest tenth of a degreedegreeAnswer: 261.9 KAnswer: 261.9 K
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.
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.
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
Movie: “The Day After Tomorrow”Movie: “The Day After Tomorrow”
Premise: global warming produces Premise: global warming produces gigantic stormsgigantic stormsIn 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!)
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 surfacePressure at 10 km Pressure at 10 km 260 hPa 260 hPaTemperature Temperature 220 K 220 KIf surface pressure If surface pressure 1000 hPa, what is 1000 hPa, what is air temperature upon reaching surface?air temperature upon reaching surface?
SolutionSolution
286.0
2601000220
hPahPaKTsfc
K323C50F122
Height Dependence of THeight Dependence of T
pcR
ppT
/
0
*
Start with
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
Step 1Step 1
Take log of both sides of *Take log of both sides of *
0
/
0
lnlnln
lnlnln
pcRp
cR
ppT
pp
cR p
Constants
Step 2: DifferentiateStep 2: Differentiate
pdzd
cRT
dzd
p
lnln
dzdp
pcR
dzdT
T p
1 **
Step 3: Hydrostatic EquationStep 3: Hydrostatic Equation
gdzdp
Substitute into **
Step 4Step 4
gpcR
dzdT
T p
1
pcg
pRT
dzdT
ExerciseExercise
Simplify the expression for dT/dz.Simplify the expression for dT/dz.
ResultResult
pcg
dzdT
Constant!
z
Parcel temperature
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
Dry-Adiabatic “Lapse Rate”Dry-Adiabatic “Lapse Rate”
p
d
pd c
RcR
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.