Lecture #4: Adiabatic Processes

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Adiabatic Processes
1000 mb
How can the first law really help me forecast thunderstorms?
Thermodynamics
M. D. Eastin
Adiabatic Processes
Outline:
 Review of The First Law of Thermodynamics
 Adiabatic Processes
 Poisson’s Relation
 Applications
 Potential Temperature
 Applications
 Dry Adiabatic Lapse Rate
 Applications
Thermodynamics
M. D. Eastin
First Law of Thermodynamics
Statement of Energy Balance / Conservation:
• Energy in = Energy out
• Heat in = Heat out
dq  cvdT  pdα
Heating
Sensible heating
Latent heating
Evaporational cooling
Radiational heating
Radiational cooling
Thermodynamics
Change in
Internal Energy
Work Done
Expansion
Compression
M. D. Eastin
Forms of the First Law of Thermodynamics
For a gas of mass m
For unit mass
dQ  dU  dW
dq  du  dw
dQ  dU  pdV
dq  du  pd
dQ  cvdT  pdV
dq  cvdT  pd
dQ  cpdT  Vdp
dq  cpdT  dp
cp  cv  nR*
cp  c v  R d
where:
p = pressure
V = volume
T = temperature
α = specific volume
U = internal energy
W = work
Q or q = heat energy
n = number of moles
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)
Thermodynamics
M. D. Eastin
Types of Processes
Isothermal 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)
Thermodynamics
M. D. Eastin
Adiabatic Processes
Basic Idea:
• No heat is added to or taken from the system
which we assume to be an air parcel
dq  cvdT  pdα  0
Parcel
dq  cpdT  dp  0
• 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
Thermodynamics
M. D. Eastin
Adiabatic Processes
P-V Diagrams:
Isobar
p
i
Isochor
Adiabat
Isotherm
f
V
Thermodynamics
M. D. Eastin
Poisson’s* Relation
A Relationship between Temperature and Pressure:
• Begin with:
cpdT  dp
• Substitute for “α” using
the Ideal Gas Law
and rearrange:
• Integrate the equation:
pα  R d T
dT R d dp

T
cp p
Tfinal

Tinitial
* NOT pronounced like “Poison”
Thermodynamics
Adiabatic Form
of the First Law
Rd
dT

T
cp
p final

p intital
dp
p
See: http://en.wikipedia.org/wiki/Simeon_Poisson
M. D. Eastin
Poisson’s Relation
A Relationship between Pressure and Temperature:
• After Integrating the equation:
Tfinal
Rd
pfinal
ln

ln
Tinitial
cp
pinitial
• After some simple algebra:
 p final 
Tfinal

 
Tinitial
 pinitial 
Rd
cp
Tfinal
 p final 

 Tinitial
 pinitial 
Rd
cp
• Relates the initial conditions of temperature and pressure to
the final temperature and pressure
Thermodynamics
M. D. Eastin
Applications of Poisson’s Relation
Tfinal
 p final 

 Tinitial
 pinitial 
Rd
cp
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?
Thermodynamics
M. D. Eastin
Applications of Poisson’s Relation
Tfinal
 p final 

 Tinitial
 pinitial 
Rd
cp
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
pfinal = 770 mb
Rd = 287 J / kg K
cp = 1004 J / kg K
Tinitial = -40ºC = 233K
Tfinal = ???
Thermodynamics
M. D. Eastin
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 cruising altitude of 30,000 feet (300 mb) is -40ºC, what is
the temperature inside the cabin?
pinitial = 300 mb
pfinal = 770 mb
Rd = 287 J / kg K
cp = 1004 J / kg K
Tinitial = -40ºC = 233K
Tfinal
 770mb
 233K

 300mb
287
1004
Tfinal  305K
Tfinal  32C
Thermodynamics
M. D. Eastin
Applications of Poisson’s Relation
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
300 mb
Tfinal
 p final 

 Tinitial
 pinitial 
Thermodynamics
Rd
cp
600 mb
-37oC
2oC
M. D. Eastin
Applications of Poisson’s Relation
Comparing Temperatures at different Altitudes:
Are they relatively warmer or cooler?
Tfinal
 p final 

 Tinitial
 pinitial 
Rd
cp
pinitial = 300 mb
pfinal = 600 mb
Tinitial = -37ºC = 236 K
Tfinal = 288 K = 15ºC
300 mb
-37oC
600 mb
2oC 15oC
Note: We could we have chosen
to expand the 600 mb parcel
to 300 mb for the comparison
Thermodynamics
M. D. Eastin
Potential Temperature
Special 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
 1000mb

θ  Tinitial
 pinitial 
 p0 
θ  T 
 p
Rd
Rd
cp
cp
1000 mb
where: p0 = 1000 mb
Thermodynamics
M. D. Eastin
Applications of Potential Temperature
Comparing Temperatures at different Altitudes:
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:
Air Parcel #2:
p = 900 mb
T = 21ºC
p = 700 mb
T = 0.6ºC
 p0 
θ  T 
 p
Rd
cp
Which parcel is relatively colder? warmer?
Thermodynamics
M. D. Eastin
Applications of Potential Temperature
Comparing Temperatures at different Altitudes:
Air Parcel #1:
p = 900 mb
T = 21ºC = 294 K
 1000mb
θ  294K

 900mb 
0.286
θ  303K
Air Parcel #2:
p = 700 mb
T = 0.6ºC = 273.6 K
 1000mb
θ  273.6K

 700mb 
0.286
θ  303K
The parcels have the same potential temperature!
Are we measuring the same air parcel at two different levels?
Thermodynamics
M. D. Eastin
Applications of Potential Temperature
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
Thermodynamics
M. D. Eastin
Applications of Potential Temperature
Potential Temperature as an Air Parcel Tracer:
• Therefore, under dry adiabatic conditions, potential
temperature can be used as a tracer of air motions
Constant θ
Thermodynamics
Constant θ
• Track air parcels moving up and down (thermals)
• Track air parcels moving horizontally (advection)
M. D. Eastin
Dry Adiabatic Lapse Rate
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):
dp
  g
dz
Hydrostatic
Relation
(more on this later)
• 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)
cpdT  dp
Thermodynamics
Adiabatic Form
of the First Law
z
Dry Adiabatic
Lapse Rate?
T
M. D. Eastin
Dry 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:
Thermodynamics
cpdT  dp
dT R d dp

T
cp p
1 dT R d 1 dp

T dz c p p dz
dp
  g
dz
 T Rd g
dT

dz
p cp
M. D. Eastin
Dry Adiabatic Lapse Rate
How does Temperature change with Height for a Rising Thermal?
• Substitute for “ρ” using
the Ideal Gas Law
and cancel terms:
 T Rd g
dT

dz
p cp
p   R dT
dT
g

dz
cp
• We have arrived at the Dry Adiabatic Lapse Rate (Γd):
dT
g
d 
 
  9.8C / km
dz
cp
Thermodynamics
M. D. Eastin
Application of the Dry Adiabatic Lapse Rate
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?
dT
  9.8C / km
dz
Tfinal  (9.8C / km) dz  Tinitial
Tfinal   9.8 * 0.8  25
Tfinal  17.2C
Mixed Layer
Constant θ
Thermodynamics
M. D. Eastin
Adiabatic Processes
Summary:
• Review of The First Law of Thermodynamics
• Adiabatic Processes
• Poisson’s Relation
• Applications
• Potential Temperature
• Applications
• Dry Adiabatic Lapse Rate
• Applications
Thermodynamics
M. D. Eastin
References
Petty, 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.
Thermodynamics
M. D. Eastin
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