Transcript Heat

Heat
First Law of Thermodynamics:


The change in internal energy of a closed
system, DU, is given by: DU  DQ  DW
where, DQ = heat added to the system,
DW = work done by the system.
Heat is one form of energy. It is energy
transferred between systems due to a change in
temperature.


Lagrangian systems: An air parcel is studied
as it moves and changes in the atmosphere.
Eulerian systems: Air at one location is
studied to see how it changes at that one
location. One must consider the transport in and
out of the air parcel.
Sensible Heat: Heat that can be sensed by
humans. It is that portion of total heat
associated with a temperature change.
Sometimes it called enthalpy (heat
function). Sensible heat per unit mass is
given by:
DQH
dQH
 Cp  DT or
 Cp  dT
mair
mair
Cp is the specific heat at constant pressure of
the material, the amount of energy necessary
to change the temperature of one unit mass of
material by 1o (usually oC or oK).
Cpd  1004.67 J kg K
For dry air:
For moist air: Cp  Cpd  1 0.84r 
where, r = mixing ratio of the air. (gvapor/gdry air)
(Sometimes given as gvapor/kgdry air)
o


For liquid water, as rain or cloud droplets,
the
o
specific heat is: Cp liquid  4200 J kg k
It varies with the temperature.
Latent Heat: Heat absorbed or released per unit
mass by a system during a change of phase.
Temperature does not measure the entire internal
energy of a substance, only the translational
kinetic energy part. Phase changes occur with no
change in temperature.
Amount of heat energy per unit mass released or
absorbed by water molecules during a phase
change is defined as: DQE
L
m wat er
Values of latent heat in text, pg. 44, are at 0oC.

Gas
Water
Ice
Process
Gains Energy Loses Energy
Evaporation
Water mol.
Environment
(vaporizaton)
Melting
Water mol.
Environment
Sublimation
Water mol.
Environment
Condensation
Environment Water
molecule Freezing Environment Water
molecule (fusion)
Deposition
Environment Water
molecule
 Specific Heat: (heat
capacity) Amount of heat
energy required to change
the temperature of one unit
mass (usually kg or gram)
of a substance by 1o
(usually oC or oK). For
water the specific heat is
not constant. Specific heat
for water varies from
4177.5 to 4216.89 J/kg oC.
Also, it varies for ice and
steam.
Lagrangian Heat Budget - Part 1
Consider only “dry” air parcels - there is no
phase changes of water occurring.
First Law of Thermodynamics: The change in
temperature results from heat energy being added
or removed from a volume of air, and/or the
volume does work - expands or contracts when
rising or sinking. Then, DQH  C  DT  DP
p
m
r
air
where,
Cp = specific heat, r = density
Cp Dt = sensible heat change, DP/r = work done
DQH
DP
 Cp  DT 
mair
r

Using the hydrostatic equation: DP  r  g  Dz
DP
then,

 g  Dz
r
DQH
 Cp  DT  g  Dz
Substituting gives:
mair
DQH
g Dz
 DT 
Rearranging gives:
m airCp
Cp
DQH
g Dz
and,
DT 

mairCp
Cp
eq. 3-5
Lapse Rate
Lapse Rate: The rate of change of temperature
with height.

Types:
• Environmental - Actual change as measured.
• Adiabatic - How a parcel of air changes as it moves
vertically in the atmosphere. No heat energy is added or
removed from the parcel. DQH  0
– Dry adiabatic change - no phase change is occurring.
– Moist adiabatic change - phase change is occurring
In the dry adiabatic process:
DQH
g Dz
g Dz
DT 

 0
mair  Cp
Cp
Cp
DT
g
9.807m s2


o
Dz
Cp
1004.67J kg  K
2
o
DT
9.807
m
s
m
kg
K
and,

 .00976 2
2
o
s

kg
m
Dz
1004.67 J kg K
s2
which gives:
o
o
DT
K 1000m
K
 .00976

 9.76
Dz
m
km
km
or:
o
DT
K
 9.76
Dry adiabatic lapse rate,  d 
Dz
9.8o K

km
km
Note: temperature decreases as parcel ascends
and increases as parcel descends.
In terms of pressure.
DQH
DP
 Cp  DT 
 0 for adiabatic process
mair
r
dP
Written in derivative form:Cp  dT   0
r
Substituting from General Gas Law
dP
P
Equation.
Cp  dT 
 Rd  T
r
Rd  T
P
dT dP Rd


T
P Cp
Integrating from P1 and T1 to P2
and T2
T2
dT Rd
T T  Cp
1
P2
dP
P P
1
Integrating from P1 and T1 to P2 and T2
T2
dT Rd
T T  Cp
1
P2
dP
P P
1
T  R
P 
2 
d
2 


ln  
ln  
T1  Cp
P1 
Since, ln x
y
  y ln x 
Rd 

T 
P C p 

2 
2 

ln 

ln

 
  
T1 
P1  


Then taking the antilog of both sides gives:
Rd
T 2  P2 Cp

 
For
dry
air,
R
/C
=
0.28571


d
p
T1  P1 
This shows that moving a parcel from a
pressure P1 to P2 will change the
temperature from T1 to a new value given
by (assuming no phase changes occur): R d
 P Cp
2 

T2  T1   
 P1 
Potential Temperature
Potential temperature: The temperature a
parcel of air would have if brought
adiabatically (no heat transfer into or out of)
from its initial state to an arbitrarily selected
height or pressure level. The standard
pressure normally used is 1000 hPa (100
kPa).
For height coordinates:
o
o
DT
K
9.8
K
remember: d 
 9.76

Dz
km
km
DT
 d
Dz
T2  T1  d Dz
Lapse rate in height units is:
This can be written as:
If we let T2 be defined as the Potential Temperature,
Q, (the temperature the parcel would have if we
moved it vertically to a particular height), then:
Q  T1   d  Dz
and,
Q  T1   d  Dz
Note: The solved problem, pg. 47, the parcel is
descending from a height of 500 m to a height of 0
meters, ground level or sea level. Dz = 500 m
Rd
Lapse rate in pressure units:
T 2  P2 Cp

 


T1  P1 
If we let T2 be the potential
temperature,
Q,
R
 P C
then:
2
d
Q  T1  
 P 

 1 
p
Typically, in pressure units, the air parcel is
moved to 1000 hPa (100 kPa).
Remember: This is for moving “dry” air
parcels, (unsaturated, no liquid water or ice
droplets), in which no energy is added or
removed to the parcel.
Remember: Virtual temperature is given by the
equation:
Tv  T1  0.6077r
If there is water vapor, liquid water droplets, the
potential temperature should take their effects
into consideration. The following virtual
potential temperature equations should be used.
Qv  Q  1 0.61 r
For non-cloudy skies:
(water vapor only)
Where r = mixing ratio. Note: must be in units of
g/g.
For cloudy skies: Qv  Q  1 0.61 rs  rL 
(liquid droplets and/or ice crystals)
where,
rs = saturation mixing ratio,
rL = mixing ratio of liquid water.
Thermodynamic Diagrams
Pressure
Temperature
Dry Adiabats
Eulerian Heat Budget
Consider a fixed volume. The change in heat of
the volume will change the temperature. That
heat change may be due to several processes: (1)
movement of air of different temperature into or
out of the volume, (2) radiation, (3) conduction
from surfaces; i.e., ground, buildings, etc., (4)
turbulence (a mixing of air). Also, internal
changes such as evaporation, condensation, (from
latent heat); chemical reactions.
The change in temperature with time can be
written in kinematic form as changes due to
fluxes moving air and due to internal changes as:
DFx DF y DFz  DSo
DT

  



Dt
Dx
Dy
Dz 
 Cp  Dt
DFx DF y DFz 




where,

 Dx
Dy
Dz 

represents flux
gradients of heat into or out of the volume (i.e.,
there is convergence into or divergence from the
volume).(Dx, Dy, Dz are lengths of sides of the
volume
DS0

represents changes in internal sources of
CP  Dt
heat.
Each of the change in flux terms is due in part to
advection, conduction, turbulence and radiation.
Advection: The process of transport of an
atmospheric property by the motion of the
atmosphere (wind).
Conduction: Transfer of energy within and
through a conductor by means of internal particle
or molecular activity.
Turbulence: A state of fluid flow in which the
instantaneous velocities exhibit irregular and
apparently random fluctuations.
Radiation: The process by which
electromagnetic radiation is propagated through
free space by virtue of joint undulatory variations
in the electric and magnetic field in space.
Advection
In the horizontal, the flux gradient due to
advection is: DFx adv  U  DT  U  D 
Dx
Dx
Dx
DFy adv
DT
D
 V
V
Dy
Dy
Dy
In the vertical, the change in temperature
due to adiabatic changes must be included.
DT

DFz adv
 W     d 
Dz

Dz
Conduction and Surface Fluxes
Conduction within a volume of air. Most heat
transfer by conduction is negligible.
Conduction at the surface of the earth is of primary
importance.
Turbulence is important in the Atmospheric
Boundary Layer (lower 1 - 2 km), and in unstable
environments.

Not important near the surface - becomes wind speed is
zero at ground level.
Effective Surface Turbulent Heat Flux
Within Atmospheric Boundary Layer effects are often
combined in an Effective Surface Turbulent Heat Flux.
Given by empirical formula for strong horizontal
advection (windy days):
FH  CH  MTsfc  Tair 
FH  CH  M sfc   air
where,
M = mean wind speed at 10 meters elevation.
CH = bulk heat transfer coefficient
(smooth surface: 2 x 10-3, rough 2 x 10-2)
Tair = temperature at 10 meters
Tsfc = temperature at ground level
During calm days with strong radiational heating, there is
more thermals occurring, greater vertical mixing. Then, a
better equation for the Effective Surface Turbulent Heat
Flux is: F  b  w     
H
H
B
sfc
ML
or:
FH  a H  w *  sfc   ML
where,
ML = potential temp. at height of 500 m
bH = convective transport coefficient = 5 x 10-4
1
aH = mixed-layer transport = 0.0063 g  z
2
i

w

  v sfc   v ML 
wB=Buoyancy velocity scale B


 v ML

w*= Deardorff velocity
13


g  z

i
w*  
 F H 


 Tv

Wind Chill
Wind Chill: The air temperature in calm
conditions that would produce the same flux
(rate of heat loss) by unclothed skin in shady
conditions that the environmental air
temperature and wind speed produce.
Note: NWS and Canada have adopted a new
Wind Chill Formula.
Wind Chill
The environmental wind speed is assumed to be at least
1.34 m/s (average walking speed = 3 mph).
Actual heat flux varies with different persons under the
same environmental conditions.
One version of the wind chill equation is:
 M  M 0.21
o 
Twind chil l  Tskin  

  Tskin  Tair 
 Mo 
As shown, there must be a difference in temperature skin to air- in order for energy to leave the skin.
Even in calm winds there should be a wind
chill value due to the movement of the
person.
New NWS formula:
WC  35.74  0.6215  T  35.75 V
0.16
  0.4275 T  V
0.16
WC in oF
T in oF
V,
wind speed, in miles/hour
Table 3-1 and Figure 3:10 are no longer
used by NWS.

Heat Index
An attempt to quantify the “sultriness” of a set of
environmental conditions and represents the extent to
which humidity aggravates the physiological effects of
high temperature on the human body.
Since, during high temperature conditions the body tries
to get rid of heat, in part, by sweating, that ability is
reduced, the internal heat of the body increases and the
person is in danger of physiological damage; such as
heat stroke, heat cramps, heat exhaustion, death.
Stull uses the 16-Element Heat Equation.
There are several equations used to calculate the Heat
Index.
 Stull’s 16-element equation:
 NWS Southern Region Technical Attachment, SR/SSD 90-23, Fort Worth, TX
from the equation by Lans P. Rothfusz
T in oF and Relative Humidity, R, in %
HI
= -42.379 + 2.04901523T + 10.14333127R - 0.22475541TR 6.83783x10 3 T 2 - 5.481717x10 -2 R 2 + 1.22874x10 -3 T 2R +
8.5282x10 -4 TR 2 1.99x10 -6 T 2 R 2
 USAToday: T in oF, Relative Humidity, RH, in %
Heat
index (HI)= - 42.379 + 2.04901523(T) + 10.14333127(RH) 0.22475541(T)(RH) - (6.83783x10-3)(T2) - (5.481717x10-2)(RH2) +
1.22874x10-3)(T2)(RH) + 8.5282x10-4)(T)(RH2) - 1.99x10-6)(T2)(RH2)
Problems
N1, N2, N3, N5(a, d, g), N6 (a, d, g), N7 (a,
e, g), N8 (a, b, c), N9 (a, b), N16 (b, d, f)
(using new NWS formula)
SHOW ALL EQUATIONS USED AND
CALCULATIONS