How can the first law really help me forecast thunderstorms?

Thermodynamics M. D. Eastin

Adiabatic Processes

How can the first law really help me forecast thunderstorms?

1000 mb


Outline:

Review of The First Law of Thermodynamics Adiabatic Processes Poisson’s Relation

Applications Potential Temperature

Applications Dry Adiabatic Lapse Rate

Applications

Adiabatic Processes


First Law of Thermodynamics

pdα dTcdq v

Statement of Energy Balance / Conservation:

• Energy in = Energy out• Heat in = Heat out

HeatingSensible heating Latent heating

Evaporational cooling Radiational heating Radiational cooling

Change in Internal Energy

Work DoneExpansion

Compression


Forms of the First Law of Thermodynamics For a gas of mass m Per unit mass

dW dUdQ pdV dUdQ

pdVdT mCdQ v

Vdp dTCmdQ p

dw du dq pd du dq

pd dTcdq v

dp dTcdq p

where: p = pressure U = internal energy n = number of molesV = volume W = work α = specific volumeT = temperature Q = heat energy m = mass

Cv = specific heat at constant volume (717 J kg-1 K-1)Cp = specific heat at constant pressure (1004 J kg-1 K-1)Rd = gas constant for dry air (287 J kg-1 K-1)R* = universal gas constant (8.3143 J K-1 mol-1)

nRCC *vp Rcc dvp


Types of ProcessesIsothermal Processes:

• Transformations at constant temperature (dT = 0)

Isochoric Processes:

• Transformations at constant volume (dV = 0 or dα = 0)

Isobaric Processes:

• Transformations at constant pressure (dp = 0)

Adiabatic processes:

• Transformations without the exchange of heat between the environment and the system (dQ = 0 or dq = 0)


Adiabatic ProcessesBasic Idea:

• No heat is added to or taken from the system which we assume to be an air parcel

• Changes in temperature result from either expansion or contraction

• Many atmospheric processes are “dry adiabatic”• We shall see that dry adiabatic process play

a large role in deep convective processes

• Vertical motions• Thermals

Parcel0pdα dTcdq v

0dp dTcdq p


Adiabatic ProcessesP-V Diagrams:

p

V

f

i

Isotherm

Isobar

Adiabat

Isochor


Poisson’s* RelationA Relationship between Temperature and Pressure:

• Begin with:

• Substitute for “α” using the Ideal Gas Law and rearrange:

• Integrate the equation:

* NOT pronounced like “Poison”

dp dTcp

See: http://en.wikipedia.org/wiki/Simeon_Poisson

TRpα d

pdp

cR

TdT

p

d

final

intital

final

initial

p

pp

dT

T pdp

cR

TdT

Adiabatic Form of the First Law

http://en.wikipedia.org/wiki/Simeon_Poisson


Poisson’s RelationA Relationship between Pressure and Temperature:

• After Integrating the equation:

• After some simple algebra:

• Relates the initial conditions of temperature and pressure to the final temperature and pressure

initial

final

p

d

initial

final

pp

cR

TT

lnln

pd

cR

initial

final

initial

final

pp

TT

pd

cR

initial

finalinitialfinal p

pT T


Applications of Poisson’s Relation

Example: Cabin Pressurization

• Most jet aircraft are pressurized to 8,000 ft (or 770 mb). If the outside air temperature at a cruising altitude of 30,000 feet (300 mb) is -40ºC, what is the temperature inside the cabin?

pd

cR

initial

finalinitialfinal p

pT T



Example: Cabin Pressurization

• Most jet aircraft are pressurized to 8,000 ft (or 770 mb). If the outside air temperature at a cruising altitude of 30,000 feet (300 mb) is -40ºC, what is the temperature inside the cabin?

pinitial = 300 mb Rd = 287 J / kg Kpfinal = 770 mb cp = 1004 J / kg K

Tinitial = -40ºC = 233KTfinal = ???

pd

cR

initial

finalinitialfinal p

pT T


Applications of Poisson’s RelationExample: Cabin Pressurization

• Most jet aircraft are pressurized to 8,000 ft (or 770 mb). If the outside air temperature at cruising altitude of 30,000 feet (300 mb) is -40ºC, what is the temperature inside the cabin?

pinitial = 300 mb Rd = 287 J / kg Kpfinal = 770 mb cp = 1004 J / kg K

Tinitial = -40ºC = 233K1004

287

final 300mb770mbK 233 T

K 305 T final

C32 T final


Comparing Temperatures at different Altitudes:

Are they relatively warmer or cooler?

• Bring the two parcels to the same level• Compress 300 mb air to 600 mb

-37oC300 mb

2oC600 mb


pd

cR

initial

finalinitialfinal p

pT T



Are they relatively warmer or cooler?

pinitial = 300 mbpfinal = 600 mbTinitial = -37ºC = 236 K

Tfinal = 288 K = 15ºC

Note: We could we have chosen to expand the 600 mb parcel to 300 mb for the comparison

-37oC300 mb

2oC600 mb


pd

cR

initial

finalinitialfinal p

pT T

15oC


Potential TemperatureSpecial form of Poisson’s Relation:

Compress all air parcels to 1000 mb• Provides a “standard”• Avoids using an arbitrary pressure level

• Define Tfinal = θ• θ is the potential temperature

where: p0 = 1000 mb

1000 mb

pd

cR

initialinitial p

1000mbT θ

pd

cR

0

ppT θ



An aircraft flies over the same location at two different altitudes and makes measurements of pressure and temperature within air parcels at each altitude:

Air parcel #1: p = 900 mbT = 21ºC

Air Parcel #2: p = 700 mbT = 0.6ºC

Which parcel is relatively colder? warmer?

Applications of Potential Temperature

pd

cR

0

ppT θ



Air Parcel #1: p = 900 mbT = 21ºC = 294 K

Air Parcel #2: p = 700 mbT = 0.6ºC = 273.6 K

The parcels have the same potential temperature! Are we measuring the same air parcel at two different levels? MAYBE


286.0

900mb1000mb294K θ

K303 θ

286.0

700mb1000mb273.6K θ

K303 θ


Potential Temperature Conservation:

• Air parcels undergoing adiabatic transformations maintain a constant potential temperature (θ)

• During adiabatic ascent (expansion) the parcel’s temperature must decrease in order to preserve the parcel’s potential temperature • During adiabatic descent (compression) the parcel’s temperature must increase in order to preserve the parcel’s potential temperature


Constant θ


Potential Temperature as an Air Parcel Tracer:

• Therefore, under dry adiabatic conditions, potential temperature can be used as a tracer of air motions

• Track air parcels moving up and down (thermals)• Track air parcels moving horizontally (advection)


Con

stan

t θ

Constant θ


How does Temperature change with Height for a Rising Thermal?• Potential temperature is a function of pressure and temperature: θ(p,T)• We know the relationship between pressure (p) and altitude (z):

• We can use this hydrostatic relation and the adiabatic form of the first law to obtain a relationship between temperature and height when potential temperature is conserved (dry adiabatic lapse rate)

Dry Adiabatic Lapse Rate

gdzdp

HydrostaticRelation

(more on this later)

dp dTcp Adiabatic Form of the First Law T

zDry Adiabatic Lapse Rate?


How does Temperature change with Height for a Rising Thermal?• Begin with the first law:

• Substitute for “α” using the Ideal Gas Law and rearrange:

• Divide each side by “dz”:

• Substitute for “dp/dz” using the hydrostatic relation and re-arrange:


gdzdp

dp dTcp

pdp

cR

TdT

p

d

dzdp

p1

cR

dzdT

T1

p

d

p

d

cg

pTR

dzdT


How does Temperature change with Height for a Rising Thermal?• Substitute for “ρ” using the Ideal Gas Law and cancel terms:

• We have arrived at the Dry Adiabatic Lapse Rate (Γd):


TRp dp

d

cg

pTR

dzdT

pcg

dzdT

kmCcg

dzdT

pd /8.9


Example: Temperature Change within a Rising Thermal• A parcel originating at the surface (z = 0 m, T = 25ºC) rises to the top of the mixed boundary layer (z = 800 m). What is the parcel’s new air temperature?

Application of the Dry Adiabatic Lapse Rate

Constant θ

Mixed Layer

kmC dzdT /8.9 initialfinal Tdz kmC( T )/8.9

258.0*8.9 T final

C17.2 T final


Summary:

• Review of The First Law of Thermodynamics• Adiabatic Processes• Poisson’s Relation

• Applications• Potential Temperature

• Applications• Dry Adiabatic Lapse Rate

• Applications

Adiabatic Processes


ReferencesPetty, G. W., 2008: A First Course in Atmospheric Thermodynamics, Sundog Publishing, 336 pp.

Tsonis, A. A., 2007: An Introduction to Atmospheric Thermodynamics, Cambridge Press, 197 pp. Wallace, J. M., and P. V. Hobbs, 1977: Atmospheric Science: An Introductory Survey, Academic Press, New York, 467 pp.

How can the first law really help me forecast thunderstorms?

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