1. World's Commission on Environment and Development Our Common Future ( The Bruntland Report ) (Oxford Univ. Press New York 1987).
2. More details on renewable energy systems are available from the National Renewable Energy Laboratory (NREL) at
3. 1 m 2 of PV panels requires 2 m 2 of available land. In 1997 the total U.S. annual electricity demand was ∼3.2 × 10 12 kWh [Energy Information Administration Report No. DOE/EIA-0383(99) (U.S. Department of Energy Washington DC 1999)]. The average solar resource per year for southwest Nevada is 2300 kWh/m 2 . If we assume 10% net plant efficiency (current technology) then solar resources per year would provide 230 kWh/m 2 . Therefore the total area needed is (3.2 × 10 12 kWh/year)/(230 kWh/m 2 per year) = 1.39 × 10 10 m 2 of collector area which requires 2.78 × 10 10 m 2 (∼10 900 square miles) of land area. A system efficiency of 15% would reduce the area to 1.9 × 10 10 m 2 (7200 square miles). Although we have used Nevada for this calculation PV panels can be placed across the entire United States. The U.S. average solar irradiance per year is 1800 kWh/m 2 . Implementation would involve 1.6- to 16-km 2 (1- to 10-square mile) "energy farms " along with the rooftops of homes and businesses and over parking lots.
4. For an estimate of the land area needed for PV panels the following information is necessary (R. L. Hulstrom personal communication). Flat-plate PV collector modules are typically placed so that they cover one-half of the available land; 1 m 2 of PV panels requires 2 m 2 of available land. In 1997 the total U.S. annual electricity demand was ∼3.2 × 10 12 kWh [Energy Information Administration Report No. DOE/EIA-0383
5. (99) (U.S. Department of Energy Washington DC 1999)]. The average solar resource per year for southwest Nevada is 2300 kWh/m 2 . If we assume 10% net plant efficiency (current technology) then solar resources per year would provide 230 kWh/m 2 . Therefore the total area needed is (3.2 × 10 12 kWh/year)/(230 kWh/m 2 per year) = 1.39 × 10 10 m 2 of collector area which requires 2.78 × 10 10 m 2 (∼10 900 square miles) of land area. A system efficiency of 15% would reduce the area to 1.9 × 10 10 m 2 (7200 square miles). Although we have used Nevada for this calculation PV panels can be placed across the entire United States. The U.S. average solar irradiance per year is 1800 kWh/m 2 . Implementation would involve 1.6- to 16-km 2 (1- to 10-square mile) "energy farms " along with the rooftops of homes and businesses and over parking lots.