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
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Take away concepts
1. Conservation of energy. 2. Black body radiation principle3. Emission wavelength and temperature (Wein’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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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
Since c is constant, frequency of EM wave emission related to electron vibrationWarm things have more energy than cold things, so ….?
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Properties of waves
Amplitude (A)Wavelength (µm)Period (sec)Frequency (1/sec)
c is constant
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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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Wein’s Law emission wavelength and temperature
max = a / T
Where:
max is wavelength of emitted
radiation (in µm) a = 2898, constant
T emitter temperature (in K)
Recall that K = T°C + 273.15
Sun’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
max = a / T
max is wavelength of emitted
radiation (in µm) a = 2898, constant
T emitter temperature (in K)
Recall that K = T°C + 273.15
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What’s your wavelength?
max = a / T(a = 2898)
Your body is 37°Cor 37+273 = 310K
max = ?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 effect of distance on radiation “the 1 / r2 rule”
Sun emission decreases in proportion to 1 / r2 of the Sun-Planet distance
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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…
Wein’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 = T4
Where:I = Black body energy radiation = (Constant) 5.67x10-8 Watts/m2/K4
T = temperature in Kelvin
= 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 sun= 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-) 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 = T4
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Solve for Teffective = 255K
Earth Effective temp: 255 K, or -18°CEarth Actual temp: 288K, or +15°C
… the difference of +33°C is due to the natural greenhouse effect.
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So what Earth’s radiation wavelength?
max = a / T
Where:
max 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