So if you had sunlight passing through the atmosphere at the equator on the a solar panel perpendicular to the direction of the light waves, how much energy in watts would it produce? I want the answer so I can speculate as to how much energy we could get out of a dyson sphere.
At the equator, the Sun provides approximately 1000 watts per square meter on the Earth’s surface. 1 kW per m² converted to one km², provides 1,000,000 kW / km² or
1,000 MW / kW² of solar energy at the equator.
Now look at the efficiency of the photovoltaic cells.
Commercially available solar panels are about 20% efficient. Technology breakthroughs may move this to more than 30% in the near future. Below is a list of photovoltaic efficiencies of the various technologies.
Efficiency Report
11.1% Thin Films: Past, Present, Future
17% Sunpower Introduces High-Efficiency Photovoltaic Modules
19% HelioVolt: Thin Film Technology Efficiency Comparison
20.5% Samsung’s Solar Cell Breakthrough
32.3% Photovoltaics: Energy for the New Millennium
72%?? Full Solar Spectrum Photovoltaic Materials Identified
Applying the efficiencies availble today provides the following energy outputs:
1,000 MW / km² x 20% efficiency = 200 MW / km²
1,000 MW / km² x 32% efficiency = 320 MW / km²
You will lose a portion of this in the energy conversion to your utilization equipment. Check out the solarexpert link below for a detailed explanation. of the factors that affect solar performance.
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The answer depends on where you are located (latitude) and the number of days without clouds. You’ve defined the location as the equator, but it will also depend on how much time has perfect conditions. Solar panels that are track the sun improve performance. Dust and dirt on the panels will greatly reduce the power output.
In southern California, the daily sun profile provides an annual average of 5.5 hours per day of peak conditions. Morning and evening condtions have low sunlight intensity.
A million square meters of solar cells. About 250,000 Watts if you get about 1 Watt per square meter from the sunlight and the cells are about 25% efficient. A single wind turbine can produce more than double this. That is why you see wind turbines but not big solar farms.
Your Dyson sphere will be well outside the orbit of Earth I presume. At double the distance from the Sun the received power at the solar cells would be a quarter of that on Earth, so a million square meters would provide about 62,500 Watts.
However the sphere would trap the entire energy output of the Sun and I don’t know what that is, you had better design some efficient heat dumping system for the outside. Otherwise things are going to get real hot real quick.
References :
At the equator, the Sun provides approximately 1000 watts per square meter on the Earth’s surface. 1 kW per m² converted to one km², provides 1,000,000 kW / km² or
1,000 MW / kW² of solar energy at the equator.
Now look at the efficiency of the photovoltaic cells.
Commercially available solar panels are about 20% efficient. Technology breakthroughs may move this to more than 30% in the near future. Below is a list of photovoltaic efficiencies of the various technologies.
Efficiency Report
11.1% Thin Films: Past, Present, Future
17% Sunpower Introduces High-Efficiency Photovoltaic Modules
19% HelioVolt: Thin Film Technology Efficiency Comparison
20.5% Samsung’s Solar Cell Breakthrough
32.3% Photovoltaics: Energy for the New Millennium
72%?? Full Solar Spectrum Photovoltaic Materials Identified
Applying the efficiencies availble today provides the following energy outputs:
1,000 MW / km² x 20% efficiency = 200 MW / km²
1,000 MW / km² x 32% efficiency = 320 MW / km²
You will lose a portion of this in the energy conversion to your utilization equipment. Check out the solarexpert link below for a detailed explanation. of the factors that affect solar performance.
–
The answer depends on where you are located (latitude) and the number of days without clouds. You’ve defined the location as the equator, but it will also depend on how much time has perfect conditions. Solar panels that are track the sun improve performance. Dust and dirt on the panels will greatly reduce the power output.
In southern California, the daily sun profile provides an annual average of 5.5 hours per day of peak conditions. Morning and evening condtions have low sunlight intensity.
References :
http://en.wikipedia.org/wiki/Solar_power
http://earthobservatory.nasa.gov/Library/Oven/
http://www.solarexpert.com/grid-tie/system-performance-factors.html
http://www.101iq.com/infosce.htm
As a rule of thumb you can reckon solar input as 1kW per square metre, so with 1 square kilometre, which is 1,000,000 square metres, the input would be 1000MW. The best PV cells are about 40% efficient, so the electrical power output would be around 400MW.
References :