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America's Solar Energy Potential Every hour, the sun radiates more energy onto the earth than the entire human population uses in one whole year. The technology required to harness the power of the sun is available now. Solar power alone could provide all of the energy Americans consume — there is no shortage of solar energy. The following paragraphs will give you the information you need to prove this to yourself and others. You do not need advanced math skills to follow and perform the arithmetic examples shown below. Anyone who can balance a checkbook or calculate the total square feet of floor space in his or her home, and understand why an area measuring 10 yards by 10 yards equals 100 square yards, can perform the following arithmetic examples and prove that American energy independence could be achieved with solar energy alone. Science tells us that every square meter of the earth's surface, when exposed to direct sunlight, receives about 1000 watts (1 kilowatt) of energy from the sun's light. Depending on the angle of sunlight, which changes with the time of day, and the geographical location [see map below], the power of the sun's light will be somewhat more or less than 1 kilowatt-hour per hour for every square meter of the earth's surface exposed to the sun.
Scientists like to measure things using the metric system. However, most Americans are unfamiliar with the metric system. (Europeans use the metric system.) It is easier for Americans to think in square feet and square yards because feet and yards are common lengths in the United States. So, for the sake of clarity and because this is written for an American audience, all measurements will be converted from meters to yards. A meter is just a little longer than a yard (about 3 and ¼ feet to a meter, compared with 3 feet to a yard). There are 10.8 square feet in a square meter. There are 9 square feet in a square yard (3x3=9). A simple calculation can accomplish the conversion from square meters to square yards. A square yard is 83.33 percent of a square meter. Prove this by multiplying 10.8 (the number of square feet in a square meter) by 83.33%. The answer is nine (the number of square feet in a square yard). If you perform the calculation you will see that the answer is slightly less than the whole number 9 (but close enough for our purpose). Using this conversion, we can say that a square yard of land in direct sunlight receives 1000 x 83.33% = 833 watts of solar energy. This calculation can also be used in reverse to convert yards to meters, simply divide by .8333 (833 divided by .8333 = 1000 rounded). Every square yard of land, if exposed to direct sunlight, receives about 833 watts of solar energy [NOTE: see the map above, and adjust the estimated amount of solar energy accordingly]. Therefore, a one square yard area exposed to continuous direct sunlight [in an optimal geographical location] for six hours will have received 6 hours x 833 watts = 4,998 watt-hours of solar energy during the course of a day. In round numbers, a one square yard area will receive about 5000 watt-hours (5 kilowatt-hours) per day of solar energy. Another way to obtain this result would be to take the 6 kilowatt-hours per meter (explained above in the third paragraph) and apply the conversion calculation (6 x 83.33% = 5 rounded). Americans can assume, at least in the Sunbelt regions of the southwestern United States, that every square yard of land exposed to direct sunlight will receive about 5 kilowatt-hours per day of solar energy.With the above information in mind, perform the following exercise: Measure an area ten yards long and ten yards wide. That would be thirty feet by thirty feet. Take a good look at the size of it. You are looking at an area covering 100 square yards. If that area were in direct sunlight all day it would receive about (5 x 100) 500 kilowatt-hours per day of solar energy. Now go look at your home electric bill. Your electric company calculates your home electric bill based on how many kilowatt-hours of electrical energy you use. Find the total amount of electricity that you have been billed for (given in kilowatt-hours). The amount of kilowatt-hours on your bill is for an entire month. If your home is a typical residential electric customer, you and your family consume between 500 and 1000 kilowatt-hours of electricity per month. Compare the quantity of electric energy your home consumed in one month with the quantity of energy the sun gives freely to a 100 square yard area exposed to direct sunlight. One hundred square yards of sunshine provides as much energy in 1 to 2 days as an average family uses in an entire month! It would be great if 100% of the sunshine became electricity, but solar energy and electricity are not the same. Technology accomplishes the conversion of solar energy to electricity. Several different technologies are used; perhaps the one that most people have heard of is the solar panel, made from photovoltaic cells called PV. For a detailed explanation of photovoltaic cells there is a very good
article on the Internet located at: Conversion of one form of energy to another always causes a loss of energy. In other words, the new form of energy will be less than the original. Efficiency is the word scientists use to describe the difference in power resulting from the conversion of one form of energy to another. The efficiency of commercially available solar panels (PV) is about 15%. This means that when a solar panel converts the sun's light to electricity, only about 15 percent of the energy in the sunlight becomes electricity. The same thing is true of gasoline in your car. Your car's engine can only convert about twenty-five percent of the energy in gasoline to mechanical energy that turns the wheels. With an average efficiency of 15 percent, a square yard of solar photovoltaic cells (PV) would produce (5 kilowatt-hours of solar energy multiplied by 15% =) .75 kilowatt-hours of electric energy per day. Solar panels (PV) covering an area ten yards by ten yards (100 square yards or 900 square feet) would produce 100 x .75 = 75 kilowatt-hours of electricity per day. Seventy-five kilowatt-hours per day is a lot of electricity for a single-family home. If part of the electricity is used to electrolyze water and produce hydrogen gas, and the gas is stored for use by a fuel cell when needed, then 100 square yards covered with solar panels would provide an average family with energy independence. Most detached family homes have more than 900 square feet of roof or that much space around their homes where solar panels could be installed. In the Southwest, if you look at any commercial or industrial park, or any typical mall or supermarket you will see that most of the buildings have flat roofs. Those roofs require insulation to lower the cost of air conditioning on hot days. If those roofs where covered with solar panels the sun would provide electricity for the air conditioning and save businesses millions of dollars per month that would otherwise be paid to the utility companies. Another technology, Concentrated Solar Power (CSP), takes a different approach to harnessing the power of the sun. Unlike photovoltaic cells, CSP uses mirrors to concentrate the sunlight on a focal point, which magnifies the suns heat. Similar to holding a magnifying glass in the sun, focusing the light onto a piece of paper until the paper catches on fire. CSP technology has more than one form. Troughs, dishes and towers are the different forms available today. A CSP dish or tower looks like a modern glass sculpture and contributes aesthetically to the landscape. CSP systems can achieve 30 percent efficiency, or about twice the efficiency of standard photovoltaic cells (2 x .75 = 1.5 kilowatt-hours per square yard per day). Large Concentrating Solar Power plants create the thermal energy equivalent to conventional fossil fuel power plants. After the sun sets, CSP plants generate electricity from cost-effective thermal storage, providing 24-hour service to the power grid. Consider the solar energy potential of one acre of land. There are 43,560 square feet in an acre. Divide the number of square feet in one acre by 9 (the number of square feet in one square yard) and you find that there are 4,840 square yards in one acre of land. A CSP dish, tower, or trough receiving an acre of sunshine would yield about (1.5 kilowatt-hours per square yard times 4,840 square yards per acre) 7,260 kilowatt-hours of electricity per day, at 30% efficiency. One acre has enough solar energy potential to yield 7.26 megawatt-hours of electricity per day, using technology that exists now. (Each thousand kilowatts is one million watts. A million watts is a megawatt.) Consider the solar energy potential of one square mile of land. A square mile is 640 acres. One square mile of sunshine has the potential of providing (640 acres x 7.26 megawatt-hours) 4,646 megawatt-hours per day of electricity using existing CSP technology at 30% efficiency. Ten thousand square miles is a plot of land 100 miles long by 100 miles wide. Multiply 640 acres by 10,000 square miles equals 6,400,000 acres. With a yield of 7.26 megawatt-hours of electricity per day per acre, a CSP system receiving 6,400,000 acres of sunshine would produce about 46,464,000 megawatt-hours of electricity per day. What does this mean? The entire State of California
uses about 50,000 megawatt-hours of electricity per hour at peak time,
and much less during off-peak hours: Sweltering California declares power emergency —Cal
ISO expects record demand at 52,336 megawatts. Suppose that California uses an average of 38,000 megawatt-hours of electricity per hour over a 24-hour period, then 24 hours x 38,000 megawatts = 912,000 megawatt-hours per day, multiplied by 365 = 333,880,000 megawatt-hours per year or 333,880 Gigawatt-Hours (GWh) per year. This supposed average is too high because in 2005, California actually consumed 288,245 Gigawatt-Hours (GWh) for the entire year: www.energy.ca.gov/electricity/gross_system_power.html A CSP farm large enough to capture the solar energy radiating on an area of land 100 miles long by 100 miles wide can produce about 50 times more electricity in a day than California consumes in a 24-hour period. For example, 50 x 912,000 = 45,600,000 megawatt-hours per day. Imagine driving your car 100 miles along one side of the CSP farm, then turn 90 degrees right and drive 100 miles along another side, then turn 90 degrees right again and drive another 100 miles, then make another 90 degree right turn and drive another 100 miles to complete driving a 100 mile square. Inside that area is 10,000 square miles or 6,400,000 acres. A 10,000 square mile solar energy farm that produces 46,464,000
megawatt-hours of electricity per day would produce 365 x 46,464,000
= 16,956,360,000 megawatt-hours of electricity per year or about
17 trillion kilowatt-hours, which is 17,000 terawatt-hours or
17 petawatt-hours. The CSP examples above assume 30 percent energy conversion efficiency and 100 percent land use. In a practical application, not all of the land area will be used. This is because of unfavorable terrain and the need for service roads and land for plant facilities. And, the solar collectors must be individually positioned for optimal orientation to the angle of sunlight and given enough space between collectors to prevent a collector from casting a shadow on adjacent collectors; the result is unused space between the collectors. For these reasons, actual electricity production will be less than the numbers shown in the examples. However, the desert regions of the southwestern United States will easily produce 7 hours of productive sunlight per day, and often exceed 1 kilowatt of solar energy per square meter, so in that respect the above calculations are conservative. All of California's electricity can be produced from 200 square miles of sunshine; 128,000 acres of desert land. Lake Mead, behind Hoover Dam, covers more than 200 square miles. Given an area the size of Lake Mead, for the production of electricity from solar energy, California would be energy independent. CSP plants seem to use a lot of land, but in reality, they use less land than hydroelectric dams for generating an equivalent electricity output, if the size of the lake behind the dam is considered. The same is true for coal plants. A CSP plant will not use any more land than a coal power plant if the amount of land required for mining and excavation of the coal is taken into consideration. If the sunshine radiating on the surface of an area 100 miles wide by 100 miles long would provide all of the electricity that America needs, every day, why would Americans hesitate to use it? There are millions of open acres in the deserts of America, where the sun?s energy does nothing more than heat rocks and sand. In 1942, General Patton established a training area in the deserts of
the southwestern United States to train and prepare American soldiers
to fight in the deserts of North Africa during World War II. Patton's
original training area was 18,000 square miles, and then expanded to 87,500
square miles (350 miles x 250 miles), an area stretching from Boulder
City, Nevada to the Mexican border and from Phoenix, Arizona to Pomona,
California. One million soldiers trained in this area using tanks, artillery
and aircraft. The desert is very resilient, there is little evidence today
of injury to the desert ecosystem. The point being, the federal government can “borrow” public land from the National and State desert Parks for the purpose of building a national solar energy system. The system would only be needed until fusion energy, or something like it, is developed, then the land would be returned to nature in the care of the public parks service. Time, sand and the desert wind would gradually remove all evidence of technologies brief occupancy. In the meantime, the lizards, turtles, snakes and scorpions would hide and sleep in the shade under the giant mirrors and troughs. The reason why solar energy has not been development on a large scale is the cost. Not the cost of sunshine, that is free. Private investors resist putting their money into solar energy projects because of the high upfront capital investment required for plant and equipment. The initial investment is what causes the price per kilowatt-hour for electricity from solar energy to be higher than the price of electricity generated from natural gas or coal. The estimated kilowatt-hour rates assigned to solar energy are not based on the cost of electricity generation, they are based on the cost of the investment capital and the requirement to earn a return on investment, or pay back the loan for the investment. Remember, the solar fuel is free. Solar energy would not be expensive if the cost of the initial capital investment is not factored into the price per kilowatt-hour. With the obvious enormous public benefit a national solar energy system would provide, why is the government holding back? Should solar energy be a public works project? We have a good example that may help answer that question. Southern California, as it is seen today, would not exist without Hoover Dam and the Colorado River Aqueduct, because without the Colorado River water the current population of Southern California would never have happened. Southern California does not have enough natural water to support the demand of a small fraction of its current population. The federal government funded Hoover Dam and the Colorado River Aqueduct. The economy of Southern California, having grown because of that funding and other public investments, has returned more in tax revenue than was spent building the dam and aqueduct, plus the sale of water and electricity has earned enough to pay the federal government back the amount of the original funding, with interest. The Following is quoted from the Executive Summary of a report by Sargent & Lundy engineering, titled: Assessment of Parabolic Trough and Power Tower Solar Technology Cost and Performance Forecasts, delivered to the U.S. DOE National Renewable Energy Laboratory:
Solar Energy News: Solar Energy Storage: High-Voltage Direct Current (HVDC) Transmission: High-Voltage Composite Electricity Transmission Lines:
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On average, and particularly in the Sunbelt regions of the Southwestern
United States, every square meter area exposed to direct sunlight will
receive about 1 kilowatt-hour per hour of solar energy. However, scientists
estimate that sunlight will provide useful solar energy for only about
6 to 7 hours per day because during the early hours and late hours of
the day the angle of the sun's light is too low. So, for example, if the
sun's light provides 6 productive hours of solar energy per day, then
a square meter of land in direct sunlight will receive about 6 kilowatt-hours
of solar energy during the course of a day.
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