Factor e Farm Energy Cycle: Difference between revisions
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*calory = 4.2 Joules | *calory = 4.2 Joules | ||
**This is used for food energy values. | **This is used for food energy values. | ||
Furthermore: | |||
*One kilowatt hour (1 kWhr) is 3,600,000 joules or 3.6 megajoules; look up Joule on wikipedia; this is also equivalent to 1 Whr = 3.6 kilojoules = 15 kilocalories = 15 kcal | |||
*Solar energy coming to the earth is about 1 kW per square meter | |||
*1 acre is about 4000 square meters | |||
*Thus, 1 acre intercepts a power of 4 megawatts | |||
*Given a single day has 6 hours of direct sunlight on average over the year, we have a total of approximately 24 megawatt hours per year coming to each acre of Factor e Farm | |||
*Multiply 1 Whr = 15 kcal by 24 million to obtain this yearly amount of calories: | |||
**15 kcal * 24 million = 360 Gcal = 86 GJ | |||
*Compare to 1 gallon of gasoline equivalent |
Revision as of 19:33, 24 March 2009
Introduction
We need to start by defining our terms. We use:
- 1 calorie = 4.2 Joules
- This is the amount of heat energy needed to raise the temperature of one gram of water by 1 degree Celsius
- 1 Calorie = 4200 J
- calorie with capital C is the heat energy needed to raise the temperature of one kilogram of water by 1 degree Celsius
- calory = 4.2 Joules
- This is used for food energy values.
Furthermore:
- One kilowatt hour (1 kWhr) is 3,600,000 joules or 3.6 megajoules; look up Joule on wikipedia; this is also equivalent to 1 Whr = 3.6 kilojoules = 15 kilocalories = 15 kcal
- Solar energy coming to the earth is about 1 kW per square meter
- 1 acre is about 4000 square meters
- Thus, 1 acre intercepts a power of 4 megawatts
- Given a single day has 6 hours of direct sunlight on average over the year, we have a total of approximately 24 megawatt hours per year coming to each acre of Factor e Farm
- Multiply 1 Whr = 15 kcal by 24 million to obtain this yearly amount of calories:
- 15 kcal * 24 million = 360 Gcal = 86 GJ
- Compare to 1 gallon of gasoline equivalent