### An Easy Way to Calculate Greenhouse Gas Reduction

Solar Panels help protect the atmosphere by generating electricity without the greenhouse gases that traditional power plants produce (you can learn more about the creation of greenhouse gas in the solar panel manufacturing process here).
The CO2 that is created by traditional power plants is measured by the government and industry in terms of metric tons / megawatt hour. A more intuitive way to think about the CO2 savings that result from your solar system is to consider the amount of atmosphere that you are

*not *polluting when the electricity you use is not coming from fossil fuel power plants.
How did we come up with the 16 city block-sized cube-of-atmosphere estimate on the Solar For Your Home page? Following are the assumptions, references and calculation that we used. You can use the same process to calculate the benefit from your own system, if you live in the Salt Lake area. Follow the references if you would like more information, or contact us. We are more than happy to help!
The size of a solar system is usually measured in DC kilowatts, its maximum output on a sunny summer day. For each DC kW installed in Salt Lake City, you should generate about 1460 AC kilowatt hours each year. A number of assumptions are used to get this number; you can learn more at:
http://www.nrel.gov/rredc/pvwatts/grid.html
To get started, CO2 and other greenhouse gases (methane and nitrous oxide) are combined into a “CO2 equivalent” number that can be measured as pounds / megawatt hour. Since sources of power vary around the country, different areas have different CO2 equivalent numbers. The number is also broken down by the time of day that the electricity is used. Since solar power is generated during “peak” hours, our peak number is about 1,340 pounds per megawatt hour, or 1.34 pounds per kilowatt hour.
The EPA knows this stuff better than anyone, and it isn’t a bad idea to spend some quality time on their website:
http://cfpub.epa.gov/egridweb/ghg.cfm
If we want to be intuitive, the weight of something that floats in the atmosphere gets us nowhere, so we need to convert to volume. One cubic foot of CO2 at sea level and at 70° F weighs .1144 pounds. We can’t be responsible for the temperature, because it tends to change now and then, but we can account for our altitude above sea level. The .1144 pounds of CO2 in a cubic foot at sea level would take up about 15% more space at the Salt Lake Airport, so we need to adjust for that.
Finally, CO2 exists in our atmosphere at

*very* low levels, so a cubic foot of pure CO2 released into the atmosphere would disperse and affect the CO2 level of a much, much larger volume of atmosphere. We are using the current 2 parts per million per year increase that is occurring in our atmosphere as the conversion factor to determine the amount of atmosphere that is being protected from change.
So, for an average-sized residential 3 kW system, the math is:
(3 kW) x (1460 kWh) x (1.34 lb/kWh) / (.1144 lb per ft³) x (1.145 altitude adjustment factor) / (2.04/1,000,000 PPM) = 28.8 billion ft³.
The cube root of this number gives you the length of each side of a cube that would contain this volume, 3,065 feet.
A Salt Lake City block is 660 feet long, so the cube of atmosphere would be the length of 4.64 blocks in each direction. This works out to a surface area for each side of the cube of 22 square blocks. Since 22 blocks in Salt Lake don’t make a square, which we need to visualize the cube, we are using the largest square of city blocks that would be covered by our cube, 4 x 4= 16.
As you can imagine, there are

*vastly* more complicated ways to calculate the problem, but those answers would be very similar to this one, and this process is easy enough that many people can work it out for their own system. Again, if you have questions, please let us know!

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