Transcript Radiation and Heat - University of Utah
Heat and Radiation
01/25/2006 CVEEN7920 Hydrometeorology Naoki Mizukami 1
Outline
1. Heat
Definition of energy and heat Conduction Advection Latent heat/Sensible heat Surface heat budget
2. 1st law of thermodynamics
Definition Thermodynamics equation Potential Temperature Lapse rate
3. Radiation
Electromagnetic wave Blackbody radiation Interaction of EM energy with materials 01/25/2006 CVEEN7920 Hydrometeorology 2
1. Heat – energy & heat
Energy: the ability to work on object measured in an unit of work [Joule = kg·m 2 ·s -2 ] 1 [J] = work done by 1 [N] of force to move 1 kg of object by 1 m distance 1 [cal] = 4.185 [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*
mv
2 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 01/25/2006 CVEEN7920 Hydrometeorology 3
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 01/25/2006 radiation convection conduction CVEEN7920 Hydrometeorology 4
1. Heat – energy & heat
Conduction (W ·m -2 )
T
x
0
q c
k
T
k
T
x
,
k
T
y
,
k
T
z
W q c >0 Where q c : heat flux [W·m -2 ] k: thermal conductivity [W·m -1 ·K -1 ] For air, k air = 0.03 [W·m -1 ·K -1 ] ) x T: Temperature [K] C 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 ] = 0.03 [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 01/25/2006 CVEEN7920 Hydrometeorology 5
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?
C 1mm W z 01/25/2006 CVEEN7920 Hydrometeorology 6
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?
C Ans. 300 [W ·m -2 ] = 0.03 [W·m -1 ·K -1 ] x ΔT[K] / 0.001[m] ΔT[K] = -10 [K] W z This temperature gradient within air mostly happens just above the ground on sunny day 01/25/2006 CVEEN7920 Hydrometeorology 7
1. Heat – energy & heat
Advection (convection) [K · s -1 ] Advection: horizontal transfer (x and y direction) Convection: vertical transfer (z direction)
T adv
V
T
u
T
x
v
T
y
w
T
z
where u, v and w: wind speed [m·s -1 ] T: Temperature [K] Warm advection -> positive Cold advection -> negative e.g. warm advection w
T
x
0 u >0 c x This is temperature change rate. Can convert heat transfer rate using specific heat 01/25/2006 CVEEN7920 Hydrometeorology 8
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?
T
y
??
c w u = ??
v = ??
Wind = 20ms-1 y x 01/25/2006 CVEEN7920 Hydrometeorology 9
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 ΔT x = 0 [ ° C/sec]
T
y
??
For y direction; v = 20*cos30 ° = 10 [ms -1 ], dT/dy = -3/50e+3[ ° C/m] So ΔT y = -10* (-5/50e+3) = 0.001[ ° C /sec] So 0.001*60*60 = 3.6 [ ° C] increase c w u = ??
Wind = 20ms-1 y v = ??
x For total, 0+ 3.6 [ ° C] = 3.6 [° C] increase 01/25/2006 CVEEN7920 Hydrometeorology 10
1. Heat – Sensible heat vs. Latent heat
Sensible 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 01/25/2006 CVEEN7920 Hydrometeorology 11
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 Water Specific heat 4186 Ice Dry air const pressure (c p ) Dry air const volume (c v ) Granite 2030 1004 717 790 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 01/25/2006 CVEEN7920 Hydrometeorology 12
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 q in sublimation q in melting q in evaporation vapor ice water q out fusion q out deposition q out condensation Latent heat (H 2 O) Phase change melting (fusion) evaporation (condensation) sublimation (deposition) Latent heat [J·kg -1 ] 3.34e+5 2.50e+6 2.83e+6 01/25/2006 CVEEN7920 Hydrometeorology 13
1. Heat – Latent heat
Computing air temperature increase due to latent heat release Assume that in dry air (mass is m air due to condensation during Δt [sec]. [kg] ),
m lquid
[kg] of liquid water is produced 1.
The amount of heat released: [J]
q
1
L v
m liquid
2. This heat raises temperature of dry air. Use dry air heat capacity, c p [J·kg -1 ·K -1 ] to convert heat [J] to temperature change [K]:
T
c p
q
1
m air
L v c p
m liquid
m air
3. Divided by Δt to find rate of temperature change
T
t
c L v p
m liquid m air
t
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1. Heat – Surface heat balance
Heat flux (W ·m -2 ) on the surface • • • • Radiation (F R ) Heat transfer into ground via conduction (F G ) Latent heat transfer (F L ) Sensible heat transfer into air via conduction and convection (F S ) Sum of all the fluxes is zero (+ : flux upward, - flux downward) F R <0 F R <0 F S >0 F R >0 F S >0 F L >0 F S <0 F L <0 F G <0 Daytime, moist surface 01/25/2006 F G <0 Daytime, dry surface CVEEN7920 Hydrometeorology F G >0 Nighttime, moist surface 15
2. 1
st
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 01/25/2006 CVEEN7920 Hydrometeorology 16
2. 1
st
law of thermodynamics - equation
Thermodynamic energy equation
c v dT dt
p
d
dt
J
Rate of change in thermal energy Heat input into the system Rate of work done by air parcel against pressure force where c v : specific heat at constant volume = 717 [J·kg -1 ·K -1 ] p: pressure [Pa] α: specific volume of dry air [m 3 ·kg -1 ] 01/25/2006 CVEEN7920 Hydrometeorology 17
2. 1
st
law of thermodynamics - equation
Entropy form of thermodynamic energy equation Equation of state (Ideal gas law)
p
d
dt
dp dt
R dT dt
Substitute into thermodynamic equation
c p dT dt
R
dp dt
J
where c p : specific heat at constant pressure = 1004 [J·kg -1 ·K -1 ] c p = c v + R R: dry air gas constant = 287 [J ·kg -1 ·K -1 ] Divided by T and again use equation of state (verify it)
c p d
ln
T dt
R
d
ln
dt p
J T
ds dt
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2. 1
st
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, p s [Pa], typically sea level pressure (~ 1000 [hPa]) From entropy form of thermodynamics equation Adiabatic -> J = 0 Rewrite in differential form
c p d
ln
T
R
d
ln
p
d
(
c p
ln
T
R
ln
p
) 0 Integrate from a state (some height) where pressure is p [Pa], and temperature is T [K], to a state of surface where pressure is p s [Pa], and temperature is Θ [K]. (verify it)
T
p s p
R c p
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2. 1
st
law of thermodynamics – potential temperature
Take logarithm of potential temperature equation and then differentiate with time (verify it)
c p d
ln
dt
J T
• 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) 01/25/2006 CVEEN7920 Hydrometeorology 20
2. 1
st
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 01/25/2006 CVEEN7920 Hydrometeorology 21
2. 1
st
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)
T
z
T
z
g c p
T
z
-> Environmental lapse rate For adiabatic process, the left side of the equation is zero
T
z
T
z
g c p
T
z
g c p
d
-> dry adiabatic lapse rate (~9.8 K/1000m) 01/25/2006 CVEEN7920 Hydrometeorology 22
Dew points Actual temperature Dry adiabatic lapse rate Moist adiabatic lapse rate 01/25/2006 CVEEN7920 Hydrometeorology 23
3. Radiation – Electromagnetic energy
Electromagnetic wave
A λ c c
c
: speed of light 3e+6 [m ·s -1 ]
λ
: Wavelength [m]
ν
: frequency [s -1 ] A: amplitude - energy is proportional to A 2 T: period [s] –time required to travel by full wavelength, ν = 1/T 01/25/2006 CVEEN7920 Hydrometeorology 24
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, E λ (T) [W ·m -2 ·μm -1 ], emitted from blackbody whose temperature is T [K] as a function of λ [ μm] ->
E
2
hc
2 5 exp
hc
1
k
T
1 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 ] 01/25/2006 CVEEN7920 Hydrometeorology 25
3. Radiation – Electromagnetic energy
Blackbody radiation curve Blackbody radiation curve 10 8 10 6 10 4 10 2 10 0 10 -2 10 -4 10 -1 01/25/2006 500 [K] 6000 [K] 250 [K] 10 0 10 1 Wavelength [ m] 10 2 CVEEN7920 Hydrometeorology 10 3 26
3. Radiation – Electromagnetic energy
Electromagnetic spectral 01/25/2006 Peak of solar radiation Peak of earth radiation CVEEN7920 Hydrometeorology 27
3. Radiation – Electromagnetic wave
Stefan-Boltzmann law: Total energy flux, E(T) [W ·m 2 ], emitted from blackbody whose absolute temperature is T [K]
E
T
4 σ: Stefan-Boltzmann constant (5.67e-8 [W·m -2 ·K -4 ]) 01/25/2006 CVEEN7920 Hydrometeorology 28
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])? 01/25/2006 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])? Ans: Write Stefan-Boltzmann equation in derivative form d
E
3
T
3 d
T
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 ] 01/25/2006 CVEEN7920 Hydrometeorology 30
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
I
E
Emissivity over all range of wavelength
I E
Infrared emissivity of some surface surface asphalt ε infrared 0.95
Snow, fresh Snow, old 0.99
0.82
sand Human skin 0.98
0.95
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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):
r
E
refrected E
incident a
E
absorbed E
incident
Transmission (transmissivity): All the coefficients depends on wavelength
t
E
transmitte d E
incident r
a
t
1 Kirchhoffs law: ε λ = a λ Objects that absorb EM energy of a particular wavelength emit EM energy of the same wavelength in the same efficiency 01/25/2006 CVEEN7920 Hydrometeorology 32
3. Radiation – interaction of radiation with materials
Absorption in atmosphere 01/25/2006 CVEEN7920 Hydrometeorology 33
Reference Texts
A first course in atmospheric Radiation, G.W. Petty Meteorology today, C.D. Ahrens Introduction to Dynamic meteorology, J.R. Holton 01/25/2006 CVEEN7920 Hydrometeorology 34
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.
1.
Drive the equation for this using rainfall rate,
P ρ
liquid [kg·m -3 ], dry air density,
ρ
air [kg·m -3 ] and
h
[ms -1 ], liquid water density, [m].
2.
Using this equation, find average latent heat heating for 5 [mm·h -1 ]. Use the following parameters h = 11 km,
ρ
air and c p = 0.689 [kg·m -3 ] ,
ρ
liquid = 1004 [J·kg -1 ·K -1 ] = 1025 [kg·m -3 ], L v = 2.5e+6 [J·kg -1 ], Hint; See slide 9 01/25/2006 CVEEN7920 Hydrometeorology 35
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 ?? wind A, 15 °C 1400 m 01/25/2006 Top: 3000 m 2000 m B, ?? °C CVEEN7920 Hydrometeorology 36
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 T air = 270 [K] (hint: use Kirchhoffs law to find emissivity of this air) 01/25/2006 CVEEN7920 Hydrometeorology 37