Solar Radiation Emission and Absorption
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Transcript of Solar Radiation Emission and Absorption
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Solar Radiation Emission and Absorption
V1003 - Climate and Society
Gisela Winckler (Lamont-Doherty Earth Observatory,
Department of Earth and Environmental Sciences)
For Peter deMenocal
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Take away concepts
1. Conservation of energy. 2. Black body radiation principle3. Emission wavelength and temperature (Wien’s Law). 4. Radiation vs. distance relation 5. Black body energy flux (Stefan-Boltzmann Law) 6. Effective temperature calculation, differences from
actual temperature.
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What is Energy?Energy: “The ability to do work”
Energy measured in Joules (1 J = 0.24 calories).
Power measured in Watts (1 J/s)
Energy is always conserved (1st law of TD).
Energy can be changed from one form to another, but it cannot be created or destroyed.
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Forms of Energy?
Thermal Energy: “The ability to do work through the release of heat (related to the temperature and phase of matter).
Electromagnetic Energy: Energy embedded in electromagnetic waves (such as light, microwaves, x-rays). When this kind of energy impinges on matter, part or all of it is converted to thermal energy
Kinetic Energy: associated with motion
Gravitational Energy
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Forms of Energy?
Thermal Energy: “The ability to do work through the release of heat (related to the temperature and phase of matter).
Electromagnetic Energy: Energy embedded in electromagnetic waves (such as light, microwaves, x-rays). When this kind of energy impinges on matter, part or all of it is converted to thermal energy
Kinetic Energy: associated with motion
Gravitational Energy
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Solar Energy
Nuclear fusion: H to HeEmits Electromagnetic
radiation (radiant E)
EM waves behave like particles and waves
EM travels at c (3 x 108 m/s)
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EM Radiation
Electromagnetic radiation is a form of energy transfer that does not require mass exchange or direct contact. The transfer can take place in a vacuum
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Electromagnetic Energy Transfer
• Radiation involves the propagation of EM waves at the speed of light c*= 3x108 m s-1. • The speed of light c*, the frequency of the EM waves u, and its wavelength l are linked through the following relationship:
c* = l u
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Properties of waves
Amplitude (A)Wavelength (µm)Period (sec)Frequency (1/sec)
c is constant
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Electromagnetic spectrum
1 µm = 1000 nm
“hot” “cold”
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Heat Energy Radiation• All bodies in the universe radiate energy at a rate which is
proportional to their absolute temperature (°Kelvin), i.e. the internal energy of the body. This is caused by molecular motion that changes based on the energy level of the molecules and atoms that the body contains.
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Blackbody Radiation
A “blackbody” absorbs and emits radiation at 100% efficiency (experimentally, they use graphite, or carbon nanotubes)
energy in = energy out
Across all wavelengths
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Planck’s Function & Blackbody Radiation
lmax σ T4
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Wien’s Law emission wavelength and temperature
lmax = a / T
Where:
lmax is wavelength of emitted
radiation (in µm) a = 2898, constant
T emitter temperature (in K)
Recall that K = T°C + 273.15Sun’s temperature is 5800KWhat’s its wavelength?
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The Sun’s temperature is 5800 K, that is the wavelength of its radiation?
a. 5000 µmb. 50 µmc. 0.5 µmd. 2 µme. 20µm
lmax = a / T
lmax is wavelength of emitted
radiation (in µm) a = 2898, constant
T emitter temperature (in K)
Recall that K = T°C + 273.15
c. 0.5 µm = 500nm
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What’s your wavelength?
lmax = a / T(a = 2898)
Your body is 37°Cor 37+273 = 310K
lmax = ?9.4 µm (far infrared)
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Earth as we see it (visible)
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Earth’s Infrared “Glow”: 15µm
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Electromagnetic spectrum
1 µm = 1000 nm
“hot” “cold”
0.5 µm
9 µm
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Visualizing emission temperatures
Blackbody applet: http://qsad.bu.edu/applets/blackbody/applet.html
Sunny day: 6000KSunset: 3200K Candlelight: 1500K
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The rate of energy flux decreases with increasing distance from the emitting object, and is directly related to the rate of increase in the surface area of the sphere. Because the surface area of a sphere increases proportional to the square of the radius we have:
S (at distance r from source) = S0 r0
2 / r2
where S is the energy flux and r is the distance from the source. In the same way:
Sr2 / Sr1 = r12 / r2
2
Where r1 and r2 are two distances along the path of the radiation from the source.
The Effect of distance on radiation“ The 1/r2 rule “
S varies as 1/r2
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EESC V2100x Fall 2010
J.E. Smerdon 22
Planetary Distances
http://www.jpl.nasa.gov/wallpaper/art/pia06890-1600-1200.jpg
Earth (~150 million km)Mars (~229 million km)
Venus (~108 million km)
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Mars is 1.52 AU
(1 AU = earth-sun distance =
1.5 x 1011 m)
Using 1/ r2 rule…
1 / (1.5*1.5) = 0.44
Mars receives ~44% of the Earth’s solar radiation.
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Jupiter is roughly 5 AU from the Sun, what fraction of Earth’s solar radiation does it get?
a. 1/2b. 1/5 c. 1/10d. 1/25e. 1/125
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Summary so far…
Wien’s Law (emission freq. and temperature)
The “1 / r2” law (radiation amt and distance)
Now let’s calculate the total radiative energy flux into or out of a planet using the:
Stefan - Boltzmann Law
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Stefan - Boltzmann LawEnergy emitted by a black body is greatly
dependent on its temperature:
I = (1-a) σ T4
Where:I = Black body energy radiation σ = (Constant) 5.67x10-8 Watts/m2/K4
T = temperature in Kelvin
a = albedo (“reflectivity”)
Example: Sun surface is 5800K, so I = 6.4 x 107 W/m2
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Calculating the Earth’s “Effective Temperature”
Easy as 1-2-3…
1. Calculate solar output.2. Calculate solar energy reaching the Earth.3. Calculate the temperature the Earth should
be with this energy receipt.
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1. Calculate solar output.
Calculate Sun temperature assuming it behaves as a blackbody (knowing that lsun= 0.5µm).
From S-B law: Isun = 6.4 x 10 7 W/m2
We need surface area of sun:Area = 4r2 = 4 (6.96x108 m) = 6.2 x 10 18 m2
Total Sun emission: 3.86 x 1026 Watts (!)SolarEmissionPower
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2. Calculate solar energy reaching the Earth.
Simple Geometry.(recall the inverse square law..)
Earth-Sun distance (D): 1.5 x 1011 mArea of sphere = 4 r 2
So, 3.86 x 10 26 Watts / (4 (1.5 x 1011 m)2 )
Earth’s incoming solar radiation: 1365 W/m2
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3. Earth energy in = energy out
You have Iearth, solve for Tearth
Stefan - Boltzmann law: Iearth = (1-a) σ Tearth4
Incoming solar radiation: 1365 W/m2
About 30% is reflected away by ice, clouds, etc.: reduced to 955 W/m2
Incoming on dayside only (DISK), but outgoing everywhere (SPHERE), so outgoing is 1/4 of incoming, or 239 W/m2
that is: (0.7)*(0.25)*1365 = energy that reaches Earth surface
Energy in = 239 W/m2 = σ T4
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Solve for Teffective = 255K
… the difference of +33°C is due to the natural greenhouse effect.
Earth Effective temp: 255 K, or -18°C
Earth Actual temp: 288K, or +15°C
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So what Earth’s radiation wavelength?
lmax = a / T
Where:
lmax is wavelength of emitted
radiation (in µm) a = 2898, constant
T emitter temperature (in K)
If Earth effective temperature is 255KWhat’s the wavelength?
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Emission Spectra: Sun and Earth
0.5 µm 15 µm9 µm
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Radiation and Matter
Also dependent upon the frequency of radiation! (next lecture)
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Emission Spectra: Sun and Earth
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Blackbody emission curves and absorption bands
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Why is the Sky blue?Rayleigh scattering of incoming,
short wavelength radiation (photons with specific energy)
Radiation scattered by O3, O2 in stratosphere (10-50 km)
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Why are sunsets red?
Blue wavelengths are scattered/absorbed
Red and orange pass through to surface