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GNwafor Group Title

What is the power density of the sun shining on a solar cell lying flat on the ground at the National Renewable Energy Laboratory in Golden, CO on September 22, 2012 (the autumnal equinox, when the center of the sun and the earth's equator are in the same plane) at noon? Assume the solar radiation to be parallel rays with a power density perpendicular to the rays of P0 = 1.35 kW/m2 and neglect losses due to atmospheric effects.

  • 2 years ago
  • 2 years ago

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  1. CreAtivSpelErr Group Title
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    The given power density is based on the flux passing through a plane that is perpendicular to it. However, your solar panel is not perpendicular to those rays, it is at an angle which means that as far as the panel is concerned, the same rays are spread out over a larger area. To solve this, find the ratio between the physical area of the solar panel and the projected area (the "flux window) that the rays actually pass through. |dw:1350601816080:dw| It will be something like: A window = tan(theta) A solar cell. Which means P window = tan(theta) P solar cell. (Note: I may have the "tan" on the wrong side.) Theta is based on the latitude of Golden, Co.

    • 2 years ago
  2. GNwafor Group Title
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    Latitude of Golden CO = 39.7556 Can be more detailed please

    • 2 years ago
  3. GNwafor Group Title
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    In the calculations involved.

    • 2 years ago
  4. abdulhadi Group Title
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    please give detail

    • one year ago
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