I have roughly 34,000 square feet of land to install parabolic trough systems. What is the peak energy output that I can expect. The systems will be used for steam generation.
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I assume your idea is solar energy heating water in troughs. There are many variables, including your latitude, weather, terrain, etc. On a clear day, on the summer solstice, you can get roughly 1,000 watts/square meter at noon. This figure is affected by latitude; the higher the latitude, the lower the angle of the sun at zenith, resulting in greater attenuation due to atmosphere.
Assuming 1,000 watts/m^2, your land, which is approximately 3,160 m^2 x 1,000 W = 3,160,000 W = 3.16 MW. This is the theoretical power of sunlight for total land utilization in ideal conditions at 100% efficiency. Assuming that you could arrange collectors to cover 3/4 of your land and your collection was about 25% efficient, you could collect:
1 kW/m^2 x 0.75 x 0.25 x 3,160 m^2 = 600 kW (during perfect weather, at noon on June 21st).
I suspect that this figure is highly optimistic however, but it still shows the potential of solar energy utilization even on your slightly larger than 3/4 acre of ground.
Assuming 1,000 watts/m^2, your land, which is approximately 3,160 m^2 x 1,000 W = 3,160,000 W = 3.16 MW. This is the theoretical power of sunlight for total land utilization in ideal conditions at 100% efficiency. Assuming that you could arrange collectors to cover 3/4 of your land and your collection was about 25% efficient, you could collect:
1 kW/m^2 x 0.75 x 0.25 x 3,160 m^2 = 600 kW (during perfect weather, at noon on June 21st).
I suspect that this figure is highly optimistic however, but it still shows the potential of solar energy utilization even on your slightly larger than 3/4 acre of ground.
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