8J. WhitePaper: GayCanough Space Solar Power vs. Terrestrial Solar Power Dr. Gay Canough - ETM Solar Works The Solar Resource The Earth intercepts 1000 W/m2 of sunlight. The US land mass alone is 1.9 billion acres or 7.689 trillion square meters. From studies of satellite photos and other measurements, the amount of solar energy (kWh or BTUs) received by the US has been figured out to be about 44,000 quadrillion (1015) BTUs or @3413 BTU/kWh per year, this converts to 12.89 quadrillion kWh. The electricity generated by all US power plants is about 1016 BTUs or 3 terawatt-hours per year. If photovoltaic cells are on average 10% efficient, then we need about 7 million acres of PV to provide all electricity or just 0.37% of the US land mass. Solar energy on Earth is an intermittent source. That 1000 W/m2 comes only on clear days between 10 AM and 2 PM Different areas of the country receive different amount of sunlight per year, depending upon the weather. For example, San Diego CA receives 6 kWh per square meter per day, while Binghamton, NY receives 3.5 kWh per square meter per day. This means that one needs twice as many solar electric panels in Binghamton as would be needed in San Diego CA for the same amount of energy. This difference in sunlight per year is only a factor of 2 and becomes less important as the cost of photovoltaics(PV) goes down. To collect 10 kWh per day in San Diego CA requires 1.66 kW of PV or 11 square meters (about 111 square ft.). In Binghamton, we need twice that area. A typical dwelling has about 1200 square ft (120 m2) of roof area, so even in Binghamton, the PV needed to run the house is only about 20% of the roof area. This is for 15% efficient crystalline silicon PV which is the standard today. Therefore, for residential power, there is no problem of space (area) needed for PV. In feet, any region which has a power density requirement of a few tens of kilowatt hours per day, the space required for PV is less than the space required for the dwellings or village buildings. Ten kWh per day will power lights and appliances (such as refrigerators, washing machines, toasters, TVs) in a typical American 3 bed-room house. The needs for dwellings in developing countries is much less at present, amounting to less than 1 kWh per day. (For example in Burundi Africa, people use about 60 Watt-hours per day, which is equivalent to running two 15 Watt contact fluorescent lights for one hour.) When megawatt-hours are needed, then the area of PV starts to become significant. From a power density/land use stand-point then, when more than 100 MWh is needed, the size of the PV array begins to be square kilometers in size. A factory needing 10 MWh could conceivably run off of PV from an area/space stand-point. A factory can easily take up a square kilometer of land. But a large city, packed into a tight space, such as New York, will not be able to run on PV without some better form of storage such as hydrogen (so that the PV could be outside the city somewhere). There are 106 Watt-hours in 1 megawatt-hour. A MWh is also equal to 1000 kWh. There are 106 square meters per square kilometer.
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