# Homework Help: Confusion on a diagram in a scientific journal

1. Aug 2, 2014

### BrainMan

1. The problem statement, all variables and given/known data
I am not sure if this goes here but I am doing research on solar panels and am confused on some of the physics in a particular scientific journal I am reading. More specifically I am having trouble with the equations on the last part of page 1 and the entire second page. I understand that they are comparing the energy of the photons and the angle of sunlight on the solar panel but I still don't know how to actually work with equations to solve for the electrical output. Also, the chart on page two is very confusing and doesn't have a very good explanation. A good explanation of these physics concepts would be greatly appreciated.

2. Relevant equations

3. The attempt at a solution

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2. Aug 2, 2014

### Bandersnatch

I'm not sure if I can spare the time at the moment to go through the second page, but the equations at the bottom of the first one are describing this effect:

That is, the more angled the surface of incidence is, the more spread out the energy.

This wiki page:
http://en.wikipedia.org/wiki/Effect_of_sun_angle_on_climate
is where the image is from. It's got a concise explanation as well.

So, using the equation (5), you can take the angle between the suface normal to solar rays and the surface of the Earth(which happens to be equal to the longitude - try using the equation for one of the poles and the equator to see the extreme cases), and the cosine of that angle will net you the fraction of the solar flux that will hit the area.
E.g., for 45 degrees, you get about 0.7(i.e., 70%) of the solar flux - 0.7*1400W/m^2=980 W/m^2.
That's how much energy hits a square metre at that lattitude.

3. Aug 2, 2014

### BrainMan

OK so is this article talking about the angle of light hitting the earth and the max amount of energy the solar panel could theoretically obtain or the angle in which the solar panel is facing the light and how much power it would be producing from the light?

4. Aug 2, 2014

### Bandersnatch

The paper talks about panels, but the principle that reduces incident energy is the same as with Earth surface.

If you manage to keep the panel facing the sun at all times, it'll always catch the full solar flux.

But the Sun is not stationary in the sky, so the angle of incident solar rays will continuously change in the absence of a tracking system, both during a day and over a year.

As the Sun travels in the sky, the angle between it's rays and the direction normal to the static solar panel's surface acts as the angle θ in fig.2. The angle is described as X in fig.3.
Once you've got X, you can tell how much solar energy is hitting the panel at a given moment.

Fig.3 describes how to get X from the instantenous position of the Sun in the sky(θe = sun elevation, and θψ = azimuth measured from the direction to the South) and the angle of elevation of the panel(θb = the angle between the surface and the direction perpendicular to the panel).

The "installation angle" later in the article is the angle between zenith and the direction perpendicular to panel's surface.

5. Aug 3, 2014

### BrainMan

So in figure 1 is that referring to the angle of the solar panel? Also, in the equations is A1 a unit of length? What is the photon density of a surface?

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Last edited: Aug 3, 2014