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.
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.
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.
Latitude of Golden CO = 39.7556
Can be more detailed please