1. CVEEN7920 Hydrometeorology
Heat and Radiation
Naoki Mizukami
01/25/2006
CVEEN7920 Hydrometeorology

2. CVEEN7920 Hydrometeorology
Outline
Heat
Definition of energy and heat
Conduction
Advection
Latent heat/Sensible heat
Surface heat budget
1st law of thermodynamics
Definition
Thermodynamics equation
Potential Temperature
Lapse rate
Radiation
Electromagnetic wave
Blackbody radiation
Interaction of EM energy with materials
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CVEEN7920 Hydrometeorology

3. CVEEN7920 Hydrometeorology
1. Heat – energy & heat
Energy: the ability to work on object
measured in an unit of work [Joule = kg·m2·s-2]
1 [J] = work done by 1 [N] of force to move 1 kg of object by 1 m distance
1 [cal] = [J]
rate of work [w =J/s]
Types of energy:
Potential energy: energy associated with position
e.g. gravitational (mgh), magnetic, electric field
Kinetic energy: energy due to motion of object
motion of body (macroscopic) -> ke = 1/2*mv2
motion of molecule (microscopic) -> Thermal energy
Heat: energy transferred between two objects (or system) associated with temperature -> unit [J]
Temperature: directly proportional to molecular or atomic motion, measure of thermal energy
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CVEEN7920 Hydrometeorology

4. CVEEN7920 Hydrometeorology
1. Heat – energy & heat
Heat transfer modes
Conduction: transfer within a substance via molecular bouncing
Advection (convection): transfer by movement of medium substance (fluid)
Radiation: transfer by electromagnetic (EM) waves
radiation
convection
conduction
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CVEEN7920 Hydrometeorology

5. CVEEN7920 Hydrometeorology
1. Heat – energy & heat
Conduction (W·m-2)
qc >0 W C
Where
qc: heat flux [W·m-2]
k: thermal conductivity [W·m-1·K-1] For air, kair = 0.03 [W·m-1·K-1] )
T: Temperature [K] x
Flux: quantity transferred into/out of unit area per unit time
e.g. energy flux (J·m-2·s-1 = W·m-2)
Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere?
Ans. 300 [W·m-2] = [W·m-1·K-1] x ΔT[K] / 0.001[m]
ΔT[K] = -10 [K]
This temperature gradient within air mostly happens just above the ground on sunny day
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CVEEN7920 Hydrometeorology

6. CVEEN7920 Hydrometeorology
1. Heat – energy & heat
Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere? z C
1mm W
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CVEEN7920 Hydrometeorology

7. CVEEN7920 Hydrometeorology
1. Heat – energy & heat
Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere? z C
Ans. 300 [W·m-2] = [W·m-1·K-1] x ΔT[K] / 0.001[m]
ΔT[K] = -10 [K] W
This temperature gradient within air mostly happens just above the ground on sunny day
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CVEEN7920 Hydrometeorology

8. CVEEN7920 Hydrometeorology
1. Heat – energy & heat
Advection (convection) [K· s-1]
Advection: horizontal transfer (x and y direction)
Convection: vertical transfer (z direction)
e.g. warm advection
where
u, v and w: wind speed [m·s-1]
T: Temperature [K] w c
Warm advection -> positive
Cold advection -> negative
u >0 x
This is temperature change rate. Can convert heat transfer rate using specific heat
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CVEEN7920 Hydrometeorology

9. CVEEN7920 Hydrometeorology
1. Heat – energy & heat
Q. Air temperature at a point 50 km north of a station is 3 ° C cooler than station. Wind is blowing 60 ° from north at 20 m s-1. What is temperature change due to advection? c
u = ??
v = ??
Wind = 20ms-1 w y x
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CVEEN7920 Hydrometeorology

10. CVEEN7920 Hydrometeorology
1. Heat – energy & heat
Q. Air temperature at a point 50 km north of a station is 5 ° C cooler than station. Wind is blowing 60 ° from north at 20 m s-1 for 1 hr. What is temperature change due to advection?
Ans.
For x direction;
u = 20*sin30° [ms-1], dT/dx = 0
so ΔTx = 0 [° C/sec] c
u = ??
v = ??
Wind = 20ms-1
For y direction;
v = 20*cos30 ° = 10 [ms-1], dT/dy = -3/50e+3[° C/m]
So ΔTy = -10* (-5/50e+3) = [° C /sec]
So 0.001*60*60 = 3.6 [° C] increase w y x
For total,
[° C] = 3.6 [° C] increase
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CVEEN7920 Hydrometeorology

11. 1. Heat – Sensible heat vs. Latent heat
Heat released or absorbed by a substance cause temperature change of a substance, not changing phase.
-> Can calculate how much heat is added or released by knowing heat capacity of substance and temperature change (next slide)
Latent heat :
Heat released or absorbed by a substance causes phase change of substance, not changing temperature.
Bowen ratio :
Ratio of energy available for sensible heat to energy available for latent heat
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CVEEN7920 Hydrometeorology

12. 1. Heat – Heat capacity and Specific heat
Heat capacity [J·K-1]: The amount of change in thermal energy due to the corresponding temperature change
Specific heat [J·kg-1·K-1]: Heat capacity per unit mass
Specific heat of some substances
Substance
Specific heat
Water
4186
Ice
2030
Dry air const pressure (cp)
1004
Dry air const volume (cv)
717
Granite
790
If you know temperature change in a system, can know how much thermal energy increases
Q. Which substance undergoes the least temperature changes given the same heat
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CVEEN7920 Hydrometeorology

13. CVEEN7920 Hydrometeorology
1. Heat – Latent heat
Latent heat [J·kg-1] : the amount of heat released or absorbed by a substance per unit mass due to its phase change
ice
water
vapor
qin
melting
qout
fusion
condensation
evaporation
sublimation
deposition
Latent heat (H2O)
Phase change
Latent heat [J·kg-1]
melting (fusion)
3.34e+5
evaporation (condensation)
2.50e+6
sublimation (deposition)
2.83e+6
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CVEEN7920 Hydrometeorology

14. CVEEN7920 Hydrometeorology
1. Heat – Latent heat
Computing air temperature increase due to latent heat release
Assume that in dry air (mass is mair [kg] ), mlquid [kg] of liquid water is produced due to condensation during Δt [sec].
The amount of heat released: [J]
2. This heat raises temperature of dry air. Use dry air heat capacity, cp [J·kg-1·K-1] to convert heat [J] to temperature change [K]:
3. Divided by Δt to find rate of temperature change
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CVEEN7920 Hydrometeorology

15. 1. Heat – Surface heat balance
Heat flux (W·m-2) on the surface
Radiation (FR)
Heat transfer into ground via conduction (FG)
Latent heat transfer (FL)
Sensible heat transfer into air via conduction and convection (FS)
Sum of all the fluxes is zero (+ : flux upward, - flux downward)
FR<0
FR<0
FR>0
FS>0
FL<0
FL>0
FS<0
FS>0
FG<0
FG<0
FG>0
Daytime, dry surface
Daytime, moist surface
Nighttime, moist surface
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CVEEN7920 Hydrometeorology

16. 2. 1st law of thermodynamics - description
Conservation of energy of system : Change in thermal energy of the system (e.g. air parcel) is equal to heat added to the system and work done by external forces.
heat
If heat is added to air parcel, the energy is used following way.
Air parcel expand (change volume), meaning air parcel does work against pressure force (external force)
Air parcel gains thermal energy inside the parcel
air parcel
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CVEEN7920 Hydrometeorology

17. 2. 1st law of thermodynamics - equation
Thermodynamic energy equation
Heat input into the system
Rate of change in thermal energy
Rate of work done by air parcel against pressure force
where
cv: specific heat at constant volume = 717 [J·kg-1·K-1]
p: pressure [Pa]
α: specific volume of dry air [m3·kg-1]
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CVEEN7920 Hydrometeorology

18. 2. 1st law of thermodynamics - equation
Entropy form of thermodynamic energy equation
Equation of state (Ideal gas law)
Substitute into thermodynamic equation
where
cp: specific heat at constant pressure = 1004 [J·kg-1·K-1]
cp = cv + R
R: dry air gas constant = 287 [J·kg-1·K-1]
Divided by T and again use equation of state (verify it)
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CVEEN7920 Hydrometeorology

19. 2. 1st law of thermodynamics – potential temperature
Definition:
Temperature, Θ [K], that a dry air parcel at pressure, p [Pa], and temperature, T [K], would have if it were moved adiabatically to a standard pressure, ps [Pa], typically sea level pressure (~ 1000 [hPa])
From entropy form of thermodynamics equation
Adiabatic -> J = 0
Rewrite in differential form
Integrate from a state (some height) where pressure is p [Pa], and temperature is T [K], to a state of surface where pressure is ps [Pa], and temperature is Θ [K]. (verify it)
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CVEEN7920 Hydrometeorology

20. 2. 1st law of thermodynamics – potential temperature
Take logarithm of potential temperature equation and then differentiate with time
(verify it)
From this equation, Adiabatic process (no heat exchange, J = 0) -> no potential temperature change
This rule is used to find dry adiabatic lapse rate (next few slides)
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CVEEN7920 Hydrometeorology

21. 2. 1st law of thermodynamics – Lapse rate
Lapse rate: the rate of air temperature decrease with height
Environmental lapse rate: actual lapse rate measured by sounding etc.
Process lapse rate: lapse rate determined by physical processes
-> Dry adiabatic lapse rate: lapse rate determined by moving dry air
parcel without any heat exchange from the environment
-> Moist adiabatic lapse rate: lapse rate determined by moving
saturated air parcel without any heat exchange from the environment,
but latent heat of condensation released
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CVEEN7920 Hydrometeorology

22. 2. 1st law of thermodynamics – Lapse rate
Take logarithm of potential temperature equation and differentiate with height, z
And then use hydrostatic equation (dp = - ρgdz) and Ideal gas law (verify)
-> Environmental lapse rate
For adiabatic process, the left side of the equation is zero
-> dry adiabatic lapse rate (~9.8 K/1000m)
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CVEEN7920 Hydrometeorology

23. CVEEN7920 Hydrometeorology
Dew points
Actual temperature
Dry adiabatic lapse rate
Moist adiabatic lapse rate
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CVEEN7920 Hydrometeorology

24. 3. Radiation – Electromagnetic energy
Electromagnetic wave c λ
c: speed of light 3e+6 [m·s-1]
λ: Wavelength [m]
ν: frequency [s-1] A
A: amplitude - energy is proportional to A2
T: period [s] –time required to travel by full wavelength, ν = 1/T
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CVEEN7920 Hydrometeorology

25. 3. Radiation – Electromagnetic energy
Any objects above zero absolute temperature ( 0K = -273 °C) emit EM energy
Blackbody: a body that emit the maximum energy it can emit or absorb given by body temperature [K]
Plank’s law: EM energy flux, Eλ(T) [W·m-2·μm-1], emitted from blackbody whose temperature is T [K] as a function of λ [μm]
->
where
h: plank’s constant 6.624e-34 [J·s]
c: speed of light 3e+6 [m·s-1]
k: Boltzmann constant 1.381e-23 [J·K-1]
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CVEEN7920 Hydrometeorology

26. 3. Radiation – Electromagnetic energy
Blackbody radiation curve
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CVEEN7920 Hydrometeorology

27. 3. Radiation – Electromagnetic energy
Electromagnetic spectral
Peak of solar radiation
Peak of earth radiation
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CVEEN7920 Hydrometeorology

28. 3. Radiation – Electromagnetic wave
Stefan-Boltzmann law:
Total energy flux, E(T) [W·m-2], emitted from blackbody whose absolute temperature is T [K]
σ: Stefan-Boltzmann constant (5.67e-8 [W·m-2·K-4])
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CVEEN7920 Hydrometeorology

29. 3. Radiation – Electromagnetic wave
Q: How does total energy flux change if temperature increases by 1 K for earth (275 [K]) and sun (6000 [K])?
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CVEEN7920 Hydrometeorology

30. 3. Radiation – Electromagnetic wave
Q: How does total energy flux change if temperature increases by 1 K for earth (275 [K]) and sun (6000 [K])?
Ans:
Write Stefan-Boltzmann equation in derivative form
For earth, dE = 3 x 5.67e-8 x 275^3 x 1 = 2.22 [W·m-2]
For sun, dE = 3 x 5.67e-8 x 6000^3 x 1 = 2.31e4 [W·m-2]
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CVEEN7920 Hydrometeorology

31. 3. Radiation – Electromagnetic wave
Blackbody emits theoretically maximum EM energy. But Actual emission, I, is less than blackbody emission.
Emissivity ε: a fraction blackbody radiation that is actual emitted depends on wavelength and substances (0 ~1)
Emissivity at a particular wavelength
Infrared emissivity of some surface
surface
εinfrared
asphalt
0.95
Snow, fresh
0.99
Snow, old
0.82
sand
0.98
Human skin
Emissivity over all range of wavelength
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CVEEN7920 Hydrometeorology

32. 3. Radiation – interaction of radiation with materials
In vacuum, characteristics of EM wave (energy) is intact.
if EM wave encounter the matter (air, water, other terrestrial materials), it is partitioned into three processes
Reflection (reflectivity):
Absorption (absorptivity):
Transmission (transmissivity):
All the coefficients depends on wavelength
Kirchhoffs law: ε λ = aλ
Objects that absorb EM energy of a particular wavelength emit EM energy of the same wavelength in the same efficiency
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CVEEN7920 Hydrometeorology

33. 3. Radiation – interaction of radiation with materials
Absorption in atmosphere
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CVEEN7920 Hydrometeorology

34. CVEEN7920 Hydrometeorology
Reference Texts
A first course in atmospheric Radiation, G.W. Petty
Meteorology today, C.D. Ahrens
Introduction to Dynamic meteorology, J.R. Holton
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CVEEN7920 Hydrometeorology

35. CVEEN7920 Hydrometeorology
Q1 Latent heat
Consider an air column of which height is h [m]. The average of condensation heating rate [K/s] over the air column is estimated by measuring rainfall rate on the ground.
Drive the equation for this using rainfall rate, P [ms-1], liquid water density, ρliquid [kg·m-3], dry air density, ρair [kg·m-3] and h [m].
Using this equation, find average latent heat heating for 5 [mm·h-1]. Use the following parameters
h = 11 km, ρair = [kg·m-3] , ρliquid = 1025 [kg·m-3], Lv = 2.5e+6 [J·kg-1], and cp = 1004 [J·kg-1·K-1]
Hint; See slide 9
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CVEEN7920 Hydrometeorology

36. CVEEN7920 Hydrometeorology
Q.2. (Lapse rate)
Wind blowing from west to east is pushing air up the mountain. On the windward side of the mountain, air temperature decrease with height at 5 °C per 1000 m. On the leeward side of the mountain, air temperature increase with decreasing height at dry adiabatic lapse rate. Surface air temperature was measured as 15 °C at A. What is the surface temperature at B ??
Top: 3000 m
wind
B, ?? °C
A, 15 °C
2000 m
1400 m
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CVEEN7920 Hydrometeorology

37. CVEEN7920 Hydrometeorology
Q.3. (Radiation)
Suppose air polluted air reflects 30% and transmits 50% of the incoming solar radiation (Assume 1368 [W·m-2] ). A) How much is absorbed by this air? B) How much is emitted by this air, assuming Tair = 270 [K] (hint: use Kirchhoffs law to find emissivity of this air)
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CVEEN7920 Hydrometeorology