Parabolic Trough Output based on Surface Area
Favorites|Homepage
Subscriptions | sitemap
HOME > > Parabolic Trough Output based on Surface Area

Parabolic Trough Output based on Surface Area

[From: ] [author: ] [Date: 11-12-05] [Hit: ]
What is the peak energy output that I can expect.The systems will be used for steam generation.-I assume your idea is solar energy heating water in troughs.There are many variables, including your latitude, weather,......
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.

-
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.

-
:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:
1
keywords: Output,Trough,Parabolic,based,on,Area,Surface,Parabolic Trough Output based on Surface Area
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .