Calculating the Sun’s Energy
The sun is a main sequence star (G-2, yellow dwarf) that is the closest stellar object to our planet. It has long been the inspiration for humans throughout civilization, as the main life-sustaining source of light and heat; without the sun’s energy, life as we know it would not exist. In this lesson, students will be able to explain the fusion reactions that are vital to the sun’s existence, measure the energy output of the sun, and compare the sun’s energy to conventional fossil-fuel resources here on earth.
Solar wind, helium, hydrogen, fusion, core, sunspot, plasma, solar eclipse, radiation
After this lesson, students should be able to:
Describe how the sun was formed, it’s current stage of stellar evolution.
Identify solar features, including sun spots, solar flares, eclipse .
Quantify the energy output of the sun, and compare/contrast solar energy to other forms used here on earth.
Introduction / Motivation
The Sun is a G2 main sequence star, and is the central feature around which our solar system is arranged. This fiery ball of hydrogen and helium is at least 4.5 billion years old, and contains over 99% of all the matter in our solar system; a million Earths could fit within the Sun! With a diameter of over 1.39 x 106 km and a mass of 1.99 x 1030 kg (330,00 Earths!), the Sun is a dynamic star with its own atmosphere that is layered with denser gasses at its core. The photosphere is the portion of the start that produces visible light, allowing us to see the radiant energy even though it lies beneath two additional atmospheric layers. Beyond the photosphere is the chromosphere, which is only visible when the photosphere is blocked, such as during a solar eclipse. The choromosphere can also be imaged using filters. The outermost region of the Sun is its’ corona, which extends many millions of kilometers beyond the Sun’s photosphere. The visible light seen in the corona is only a fraction of what is emitted from the photosphere, though the corona often shows up brilliantly during a solar eclipse (see figure 22.2). It appears as star-burst shaped spicules that flare-out from the Sun in all directions. Some of this energy can escape the Sun’s atmosphere, flowing into space as in streamers of protons and electrons, known as solar wind.
The Sun’s corona peeks out during the 1998 total solar eclipse in Antigua, West Indies. This is a phenomenon known as “the diamond ring” effect, which takes place just seconds before totality. Photo credit: Kelly Knight
Units of Measurement
Distances are measured in light-years (ly), or the AU (astronomical unit: the distance from the Earth to the Sun). Energy units include BTU’s, kilowatt-hours.
For a video of the Sun’s layers and images of solar flares, click here:
Visible light is part of the electromagnetic (EM) radiation (EMR) that is emitted by objects. The wavelengths of radiation emitted include dangerous cosmic rays, X-rays, radio waves, infra-red and ultra-violet radiation. Visible light is a small part of the full spectrum of EMR. For our sun, the EM radiation is created by processes such as hydrogen fusion—a thermonuclear reaction. In nebulae, gasses are heated enough to incandesce (glow) much the same as a fluorescent light bulb does. The sources of heat for glowing nebula may be EM radiation from nearby stars or from hear generated by compression of these same gasses.
|Astronomical Unit (AU)||The average distance between the Earth and the Sun: 1.5 x 108 km = 93 million miles.|
|Chromosphere||The layer in the solar atmosphere between the photosphere and corona.|
|Corona||The sun’s outermost atmosphere.|
|Hydrogen fusion||The nuclear-reaction that fuses hydrogen atoms, producing heat and Helium.|
|Luminosity||The electromagnetic radiation that is emitted from a star or other stellar object. Sometimes expressed as a flux, or amount per unit area.|
|Photosphere||The portion of the sun’s atmosphere where visible light as at is maxiumum.|
|Plasma||A hot, ionized gas.|
|Solar wind||The expulsion of electrons and protons from the sun; occurs in a radial direction.|
|Sunspot||Isolated ‘cool spots’ in the sun’s photosphere, caused by protrusionsof the the sun’s magnetic field.|
Watt’s Better? Solar Energy or Fossil Fuels?: This exercise will walk students through basic energy calculations that investigate the different energy values from solar radiation and the burning of fossil fuels.
Conversion Chart for Energy Units
One kilowattt-hour = 3,413 Btu One barrel of oil = 1,640.8 kilowatt-hours
1,367 W per square meter = solar constant Solar radiation (entire sun) = 3.83 x 1023 kW
The sun’s surface is an amazing 5700 K! Through super-heated hydrogen and helium, large amount of energy are irradiated. The solar radiation for the entire sun is 3.83 x 1023 kW (Stanford Solar Center, 2016). For comparison, imagine the radiant energy from a standard household light bulb (100 W). As the sun’s radiation travels away from the sun and towards the earth, the rays become more diffuse. To account for this “loss” in solar energy, scientists report incident radiation in terms of the amount of incoming energy that would strike a plane on the Earth’s surface at 90 degrees (perpendicular) to the incoming angle of the sun’s rays (see illustration below). This is the number is called the ‘solar constant’, and is reported in energy per area; the solar constant is 1,367 W per square meter.
Incident solar energy hitting the earth’s surface (ITACA, 2016)
The Earth, however, is a sphere, which must be accounted for in our calculations. The formula for calculating the radiation incident on a spherical surface area is, (where the radius of the Earth, R, is 6,378 km):
solar constant x πR2 =1367 W/m2 x πR2
This value is then divided by half, to account for the illuminated side of the earth (the opposing, or “dark side”, receives minimal incoming energy). This value, 684 W/m2 is the average amount of energy incident on the side of the earth that is facing the sun.
This incoming energy, however, is not completely received by Earth. The Earth’s atmosphere absorbs and diffuses incoming radiation, and only about 30% makes it to the Earth’s surface. This means the available energy for capture is:
0.7 * 684 W/m2 = 479 W/m2
If we (generously) assume that the amount of daylight during a typical day is 12 hours, we can calculate the amount of energy harnessed per day:
479 W/m2 * 12 hours = 5748 Wh/m2day
The energy E in kilowatt-hour (kWh) is equal to the power P in watts (W), times the time period t in hours (hr) divided by 1000:
E(kWh) = P(W) × t(hr) / 1000
So kilowatt-hour = watt × hour / 1000
or 5748 Wh/m2day or 5.75 kWh/m2day
What does this mean? A 1 m x 1 m solar panel would be able to receive 5.75 kWh of energy per day. For comparison, the average US consumer uses 911 kWh per month, or 30 kWh per day.
1. How many solar panels would it take to power a residential property, assuming (generously) that a solar panel allows for 100% conversion? Show your work.
Now let’s look at fossil fuels…review the energy bill for your home (or use one of the sample energy bills provided), and write-down how many kWh of energy you used for the current month.
2. How many kWh did you use in your home?
3. How many BTU’s were required to create the kilowatt hours your consumed? How many barrels of oil does this amount to? Show your work
4. What is the going rate for a barrel of oil?
6. Compare and contrast using solar power vs. fossil fuels? What are the benefits of each? What are some drawbacks?
Appendix: Sample electricity bill