Transcribed

## Transparency: What Is the Easiest Way to Power a Lightbulb

HOW MUCH FUEL DOES IT TAKE TO POWER A LIGHTBULB FOR A YEAR? HOW MUCH ENERGY-WHETHER ELECTRIC, COAL, NUCLEAR, OR OTHERWISE-IS REQUIRED FORA 100-WATT LIGHTBULB TO RUN FOR A YEAR, 24 HOURS A DAY? Kilowatt-hour (kWh) = unit of energy equal to 1.000 watt hours A 100-watt light uses 0.1 kilowatt-hours worth of energy in an hour How much energy does an electric 100-watt lightbulb use a year? 0.1 kW x 8,760 hours in a year = 876 kWh WIND COAL O 1.5 megawatt turbine x 365 days x 24 hours x 25% capacity = 3,285 mWh = 3,285,000 kWh * 876 kWh 3,285.000 kWh/year = 2.67 x 104 years O 2.67 x 10 years = 2 hours 20 minutes 9 seconds 1.5 MW turbine operating at 25% capacity * Burning 1 ton of coal creates 2,460 kWh of energy. 714 pounds for 2 hrs, 20 min 9 sec. O 876 kWh-2,460 kWh/ton = 0.357 tons BURNED COAL *0.357 tons = 714 pounds SOLAR NUCLEAR 8 days 18 hrs 14 min 24 sec. on 100 square meters of solar panels * Assuming a yield between 20-25% (average solar convertion efficiency), .035 pounds O 1 pound of the uranium 235 isotope can produce 10,246,469 kWh of energy a one square meter solar panel can produce 1 kWh of energy in a day O 876 kWh=10,246,469 kWh/pound = 8.5 x 10pounds O 100 one square meter panels producing 100 kWh would require 8 days 18 hours 14 minutes and 24 seconds NATURAL URANIUM O Divide this by .35 (the average thermal efficiency of uranium 235 is only 35%) = 2.4 x 10pounds O Divide this by.007 (uranium 235 isotope needed for nuclear fission makes up only .7% of raw uranium) = .035poundsf natural uranium. HYDROELECTRIC NATURAL GAS O The power of a hydro-electric turbine = (height of dam) x (river flow) x (efficiency) + 11.8 (converts units of feet and ++++H seconds into kilowatts) * Assuming: dam height(10 feet) x river flow(500 ft per second) x (80% efficiency) - 11.8 = 339 kW turbine 339 kW turbine operating at 80% efficiency O 876 kWh + 339 kW = for 2 hrs and 35 min. 2 hours and 35 minutes 143 pounds O 1 pound of natural gas can produce 6.12 kWh of energy O 876 kWh + 6.12 kWh/pound = 143 pounds of natural gas with 500 ft' of water per second falling from 10 ft. NATURAL GAS HOW MUCH FUEL DOES IT TAKE TO POWER A LIGHTBULB FOR A YEAR? HOW MUCH ENERGY-WHETHER ELECTRIC, COAL, NUCLEAR, OR OTHERWISE-IS REQUIRED FORA 100-WATT LIGHTBULB TO RUN FOR A YEAR, 24 HOURS A DAY? Kilowatt-hour (kWh) = unit of energy equal to 1.000 watt hours A 100-watt light uses 0.1 kilowatt-hours worth of energy in an hour How much energy does an electric 100-watt lightbulb use a year? 0.1 kW x 8,760 hours in a year = 876 kWh WIND COAL O 1.5 megawatt turbine x 365 days x 24 hours x 25% capacity = 3,285 mWh = 3,285,000 kWh * 876 kWh 3,285.000 kWh/year = 2.67 x 104 years O 2.67 x 10 years = 2 hours 20 minutes 9 seconds 1.5 MW turbine operating at 25% capacity * Burning 1 ton of coal creates 2,460 kWh of energy. 714 pounds for 2 hrs, 20 min 9 sec. O 876 kWh-2,460 kWh/ton = 0.357 tons BURNED COAL *0.357 tons = 714 pounds SOLAR NUCLEAR 8 days 18 hrs 14 min 24 sec. on 100 square meters of solar panels * Assuming a yield between 20-25% (average solar convertion efficiency), .035 pounds O 1 pound of the uranium 235 isotope can produce 10,246,469 kWh of energy a one square meter solar panel can produce 1 kWh of energy in a day O 876 kWh=10,246,469 kWh/pound = 8.5 x 10pounds O 100 one square meter panels producing 100 kWh would require 8 days 18 hours 14 minutes and 24 seconds NATURAL URANIUM O Divide this by .35 (the average thermal efficiency of uranium 235 is only 35%) = 2.4 x 10pounds O Divide this by.007 (uranium 235 isotope needed for nuclear fission makes up only .7% of raw uranium) = .035poundsf natural uranium. HYDROELECTRIC NATURAL GAS O The power of a hydro-electric turbine = (height of dam) x (river flow) x (efficiency) + 11.8 (converts units of feet and ++++H seconds into kilowatts) * Assuming: dam height(10 feet) x river flow(500 ft per second) x (80% efficiency) - 11.8 = 339 kW turbine 339 kW turbine operating at 80% efficiency O 876 kWh + 339 kW = for 2 hrs and 35 min. 2 hours and 35 minutes 143 pounds O 1 pound of natural gas can produce 6.12 kWh of energy O 876 kWh + 6.12 kWh/pound = 143 pounds of natural gas with 500 ft' of water per second falling from 10 ft. NATURAL GAS HOW MUCH FUEL DOES IT TAKE TO POWER A LIGHTBULB FOR A YEAR? HOW MUCH ENERGY-WHETHER ELECTRIC, COAL, NUCLEAR, OR OTHERWISE-IS REQUIRED FORA 100-WATT LIGHTBULB TO RUN FOR A YEAR, 24 HOURS A DAY? Kilowatt-hour (kWh) = unit of energy equal to 1.000 watt hours A 100-watt light uses 0.1 kilowatt-hours worth of energy in an hour How much energy does an electric 100-watt lightbulb use a year? 0.1 kW x 8,760 hours in a year = 876 kWh WIND COAL O 1.5 megawatt turbine x 365 days x 24 hours x 25% capacity = 3,285 mWh = 3,285,000 kWh * 876 kWh 3,285.000 kWh/year = 2.67 x 104 years O 2.67 x 10 years = 2 hours 20 minutes 9 seconds 1.5 MW turbine operating at 25% capacity * Burning 1 ton of coal creates 2,460 kWh of energy. 714 pounds for 2 hrs, 20 min 9 sec. O 876 kWh-2,460 kWh/ton = 0.357 tons BURNED COAL *0.357 tons = 714 pounds SOLAR NUCLEAR 8 days 18 hrs 14 min 24 sec. on 100 square meters of solar panels * Assuming a yield between 20-25% (average solar convertion efficiency), .035 pounds O 1 pound of the uranium 235 isotope can produce 10,246,469 kWh of energy a one square meter solar panel can produce 1 kWh of energy in a day O 876 kWh=10,246,469 kWh/pound = 8.5 x 10pounds O 100 one square meter panels producing 100 kWh would require 8 days 18 hours 14 minutes and 24 seconds NATURAL URANIUM O Divide this by .35 (the average thermal efficiency of uranium 235 is only 35%) = 2.4 x 10pounds O Divide this by.007 (uranium 235 isotope needed for nuclear fission makes up only .7% of raw uranium) = .035poundsf natural uranium. HYDROELECTRIC NATURAL GAS O The power of a hydro-electric turbine = (height of dam) x (river flow) x (efficiency) + 11.8 (converts units of feet and ++++H seconds into kilowatts) * Assuming: dam height(10 feet) x river flow(500 ft per second) x (80% efficiency) - 11.8 = 339 kW turbine 339 kW turbine operating at 80% efficiency O 876 kWh + 339 kW = for 2 hrs and 35 min. 2 hours and 35 minutes 143 pounds O 1 pound of natural gas can produce 6.12 kWh of energy O 876 kWh + 6.12 kWh/pound = 143 pounds of natural gas with 500 ft' of water per second falling from 10 ft. NATURAL GAS HOW MUCH FUEL DOES IT TAKE TO POWER A LIGHTBULB FOR A YEAR? HOW MUCH ENERGY-WHETHER ELECTRIC, COAL, NUCLEAR, OR OTHERWISE-IS REQUIRED FORA 100-WATT LIGHTBULB TO RUN FOR A YEAR, 24 HOURS A DAY? Kilowatt-hour (kWh) = unit of energy equal to 1.000 watt hours A 100-watt light uses 0.1 kilowatt-hours worth of energy in an hour How much energy does an electric 100-watt lightbulb use a year? 0.1 kW x 8,760 hours in a year = 876 kWh WIND COAL O 1.5 megawatt turbine x 365 days x 24 hours x 25% capacity = 3,285 mWh = 3,285,000 kWh * 876 kWh 3,285.000 kWh/year = 2.67 x 104 years O 2.67 x 10 years = 2 hours 20 minutes 9 seconds 1.5 MW turbine operating at 25% capacity * Burning 1 ton of coal creates 2,460 kWh of energy. 714 pounds for 2 hrs, 20 min 9 sec. O 876 kWh-2,460 kWh/ton = 0.357 tons BURNED COAL *0.357 tons = 714 pounds SOLAR NUCLEAR 8 days 18 hrs 14 min 24 sec. on 100 square meters of solar panels * Assuming a yield between 20-25% (average solar convertion efficiency), .035 pounds O 1 pound of the uranium 235 isotope can produce 10,246,469 kWh of energy a one square meter solar panel can produce 1 kWh of energy in a day O 876 kWh=10,246,469 kWh/pound = 8.5 x 10pounds O 100 one square meter panels producing 100 kWh would require 8 days 18 hours 14 minutes and 24 seconds NATURAL URANIUM O Divide this by .35 (the average thermal efficiency of uranium 235 is only 35%) = 2.4 x 10pounds O Divide this by.007 (uranium 235 isotope needed for nuclear fission makes up only .7% of raw uranium) = .035poundsf natural uranium. HYDROELECTRIC NATURAL GAS O The power of a hydro-electric turbine = (height of dam) x (river flow) x (efficiency) + 11.8 (converts units of feet and ++++H seconds into kilowatts) * Assuming: dam height(10 feet) x river flow(500 ft per second) x (80% efficiency) - 11.8 = 339 kW turbine 339 kW turbine operating at 80% efficiency O 876 kWh + 339 kW = for 2 hrs and 35 min. 2 hours and 35 minutes 143 pounds O 1 pound of natural gas can produce 6.12 kWh of energy O 876 kWh + 6.12 kWh/pound = 143 pounds of natural gas with 500 ft' of water per second falling from 10 ft. NATURAL GAS HOW MUCH FUEL DOES IT TAKE TO POWER A LIGHTBULB FOR A YEAR? HOW MUCH ENERGY-WHETHER ELECTRIC, COAL, NUCLEAR, OR OTHERWISE-IS REQUIRED FORA 100-WATT LIGHTBULB TO RUN FOR A YEAR, 24 HOURS A DAY? Kilowatt-hour (kWh) = unit of energy equal to 1.000 watt hours A 100-watt light uses 0.1 kilowatt-hours worth of energy in an hour How much energy does an electric 100-watt lightbulb use a year? 0.1 kW x 8,760 hours in a year = 876 kWh WIND COAL O 1.5 megawatt turbine x 365 days x 24 hours x 25% capacity = 3,285 mWh = 3,285,000 kWh * 876 kWh 3,285.000 kWh/year = 2.67 x 104 years O 2.67 x 10 years = 2 hours 20 minutes 9 seconds 1.5 MW turbine operating at 25% capacity * Burning 1 ton of coal creates 2,460 kWh of energy. 714 pounds for 2 hrs, 20 min 9 sec. O 876 kWh-2,460 kWh/ton = 0.357 tons BURNED COAL *0.357 tons = 714 pounds SOLAR NUCLEAR 8 days 18 hrs 14 min 24 sec. on 100 square meters of solar panels * Assuming a yield between 20-25% (average solar convertion efficiency), .035 pounds O 1 pound of the uranium 235 isotope can produce 10,246,469 kWh of energy a one square meter solar panel can produce 1 kWh of energy in a day O 876 kWh=10,246,469 kWh/pound = 8.5 x 10pounds O 100 one square meter panels producing 100 kWh would require 8 days 18 hours 14 minutes and 24 seconds NATURAL URANIUM O Divide this by .35 (the average thermal efficiency of uranium 235 is only 35%) = 2.4 x 10pounds O Divide this by.007 (uranium 235 isotope needed for nuclear fission makes up only .7% of raw uranium) = .035poundsf natural uranium. HYDROELECTRIC NATURAL GAS O The power of a hydro-electric turbine = (height of dam) x (river flow) x (efficiency) + 11.8 (converts units of feet and ++++H seconds into kilowatts) * Assuming: dam height(10 feet) x river flow(500 ft per second) x (80% efficiency) - 11.8 = 339 kW turbine 339 kW turbine operating at 80% efficiency O 876 kWh + 339 kW = for 2 hrs and 35 min. 2 hours and 35 minutes 143 pounds O 1 pound of natural gas can produce 6.12 kWh of energy O 876 kWh + 6.12 kWh/pound = 143 pounds of natural gas with 500 ft' of water per second falling from 10 ft. NATURAL GAS

# Transparency: What Is the Easiest Way to Power a Lightbulb

shared by rmmojado on Dec 28
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How much energy—whether electric, coal, nuclear, or otherwise—is required for a 100-watt lightbulb to run for a year, 24 hours a day? See the answer in our latest infographic. A collaboration b...

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