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.