Optimal Panel Elevation Angle For a Single Day

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wwhitney

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More a math/physics question, but say you're setting up a portable, adjustable ground mount PV system before dawn for a single day event. So you just want to maximize your solar production over the course of that day, which will be clear all day. Pointing your array due south is clearly optimal, but what is the optimal elevation angle?

A lot of discussions seem to assume that the sun's elevation angle at solar noon is the correct answer, but that can't be exactly true. A little lower has to be better, but is it on the order of 0.5 degrees, or 5 degrees? This must be a solved problem, so I was hoping someone could point me to the solution and save me from looking up the formula for the sun's angle over the course of the day and doing the calculus.

Cheers, Wayne
 
It depends on the latitude and time of year. The sun moves north and south in the sky over the course of the year, so the optimal elevation angle will change. If you are on the equator, the sun will be in the northern sky half of the year, in the southern sky half of the year, and directly overhead for two days. Look at a sun path diagram for your latitude for an illustration. To maximize the output in that plane over a single day, you would want to be halfway between the elevation at sunrise and the elevation at solar noon. But it would also be influenced by the incident angle of the sun throughout the day as it moves from east to west, so it might not be quite that simple.
 
d0nut said:
To maximize the output in that plane over a single day, you would want to be halfway between the elevation at sunrise and the elevation at solar noon. But it would also be influenced by the incident angle of the sun throughout the day as it moves from east to west, so it might not be quite that simple.
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Definitely not that simple. If the sun's azimuth was constant (due south all day, only the elevation changes), and if it moved up in the sky at constant angular velocity, then reversed at the same velocity, half the noon elevation would be the answer (since the elevation at sunrise is 0).

But neither of those assumptions is true. I believe the elevation angle is changing more rapidly at sunrise and sunset than around solar noon; and since the sun's azimuth is changing, the elevation mismatch around sunrise and sunset has less impact on production that it does around solar noon. Both of those effects mean that the optimal elevation will be much closer to the noon elevation than to half the noon elevation.

Cheers, Wayne
 
The sun does move through the sky at the same angular velocity. That is why if you have a sundial with its gnomon aligned with the celestial poles, the hour lines will be evenly spaced on the face of the dial. The shadow will revolve on the face of the dial at a constant rate. The sun's altitude will be at 0 at sunrise, but the azimuth will depend on your latitude and the time of year.

If you are on the equator and we call east 0, the only time the sun would rise exactly at 0 would be during the vernal equinox and the autumnal equinox. From March 21 to September 21, the sunrise would move from 0 to -23.44 back to 0, and then from September 21 to March 21 the sunrise would move from 0 to 23.44 back to 0.

The light incident on the solar panel would vary throughout the day, proportional to the cosine of the angle between the direction of the sun and the surface normal of your panel. That is the portion you would have to maximize to get the most output out of your panels.
 
The more I think about it, I think solar noon might be the correct answer. If you are oriented in the north/south direction, at solar noon you would have the maximum production possible out of your panels since the incident energy*cos(0deg) would be maximized. If you move off of solar noon, you will never get the maximum output out of the panel since you will always be multiplying the incident energy by cos(>0deg), giving a value less than 1.
 
d0nut said:
The sun does move through the sky at the same angular velocity.
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That's true for some notion of angle, but not for elevation angle, which I was referring to.

d0nut said:
The sun's altitude will be at 0 at sunrise, but the azimuth will depend on your latitude and the time of year.
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Yes, if I'm understanding correctly, then for latitude Phi, azimuth A (with north = 0 degrees), and solar declination delta (0 at the equinox, 23.4 at the summer solstice), the controlling equation is cos Phi * cos A = sin delta.

Cheers, Wayne
 
It gets even more complicated because you would have to compare the incident energy normal to your panel to the output curve of your PV panel and the output changes in the panel as it would heat up. We would have to maximize both the sun angle and the PV output curve and minimize the degradation of the output due to high temperatures.

No wonder everyone just says to put them at solar noon. You can quickly turn this into a very difficult problem to maximize across all of the variables. We also may be measuring with a micrometer and cutting with a chainsaw at this point.
 
d0nut said:
The more I think about it, I think solar noon might be the correct answer.
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Here's my argument that the optimum has to be a non-zero amount below the sun's elevation at solar noon:

The instantaneous production of the panel is proportional to the sun's unit vector in the sky dot product with the panel's unit normal. And the day's production is proportional to the integral of that over the day. If the panel normal = sun vector at noon, that dot product will be 1, but the derivative of the dot product with respect to panel elevation will be 0. So lowering the panel elevation a little below the sun's noon elevation will have only a second order negative effect on production at times near noon. But it will have a first order positive effect at times away from noon. That means the net effect will be to increase production.

Cheers, Wayne
 
d0nut said:
It gets even more complicated because you would have to compare the incident energy normal to your panel to the output curve of your PV panel and the output changes in the panel as it would heat up.
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For this question I'm happy to stick to the geometry and just model the panel output as proportional to irradiance. I agree than modeling panel temperature and panel/system efficiency at different irradiances would be complicated.

Cheers, Wayne
 
Remember to account for the difference between true (solar) south and magnetic south if you use a magnetic compass.
There's a fairly significant difference in northern California.
 
wwhitney said:
Here's my argument that the optimum has to be a non-zero amount below the sun's elevation at solar noon:
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Here's a simpler way to state the same argument:

For the given day's sun path, consider the total production P as a function of the panel elevation angle E. For angles near the optimum P will be a smooth function of E. Therefore at the optimal angle E, P'(E) = 0. But when E = the sun's noon elevation, it's clear that increasing E will decrease P (since at all times of the day, increasing E will increase the angle between the panel and the sun). So at that elevation, P' is non-zero, and that is not the optimum elevation angle.

Cheers, Wayne
 
Output of a PV module is even more complex than discussed above. There are 3 major components of the sunshine for a clear sky: direct beam, diffuse, and reflected. Each of these varies by location, time of year, and other factors such as water vapor in the air (affects the spectrum). Depending on the cell technology, there are differences in spectral response. Ambient temperature varies during the day, warmer in the afternoon.

There are some very complex computer programs that can calculate the effect of these parameters and estimate the performance of a PV module in specific situations. No simple rules apply for short cuts.

I have an older version of one of these programs, based on NREL formulas of 30 years ago that have proven reasonably accurate. If one uses such a program and varies the azmiuth E-W and calculates daily energy output, you will see that an orientation slightly to the East produces more energy than directly south because of the temperature effects. Like wise, an elevation angle that is less than the noon solar elevation will produce more daily energy. The same procedure can be used to calculate annual or monthly energy, provided that local solar radiation data is available.

There are many other factors (such as anti-reflective coatings) to consider if you want to be more accurate. Modules degrade in operation, new vs. old.

The price of PV these days makes accuracy less important, just use more PV to be safe. There is lots of continuing work in this area to support financing of the large PV projects.
 
wwhitney said:
For this question I'm happy to stick to the geometry and just model the panel output as proportional to irradiance. I agree than modeling panel temperature and panel/system efficiency at different irradiances would be complicated.
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OK, leaving aside the complexities others have noted, I figured out the geometric portion of the problem. This is for a model where production is proportional to the dot product of the sun's direction with the panel normal vector. The dot product is bilinear, so you can pull the dot product out of the integral.

That means for the time period under consideration you can just express the sun's position as a function of time as points on the unit sphere, and then integrate each coordinate separately over time. Then the panel normal should be parallel to the resulting vector to maximum the production. If you have an efficiency forecast as a function of time (shading, heating, etc) you can just multiply the coordinates by the efficiency before integrating. And if you want to maximize economic output, you can multiply the coordinates by an economic factor before integrating.

Cheers, Wayne
 
wwhitney said:
That means for the time period under consideration you can just express the sun's position as a function of time as points on the unit sphere, and then integrate each coordinate separately over time. Then the panel normal should be parallel to the resulting vector to maximum the production. If you have an efficiency forecast as a function of time (shading, heating, etc) you can just multiply the coordinates by the efficiency before integrating. And if you want to maximize economic output, you can multiply the coordinates by an economic factor before integrating.
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Can you translate that to simple(r) English, please? :unsure:
 
LarryFine said:
Can you translate that to simple(r) English, please? :unsure:
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At any given instant, the solar panel's output will be maximized by pointing it directly at the sun. For a period of time (an hour, a day, a month, a year), if you want to maximize production with a single panel orientation, you need to average of the sun's position in the sky over the time period, and point the panel in that average direction.

Which raises the question of how you average directions in the sky. The meaning here, for two points, is just to draw the line across the sky connecting those two points and take the half-way point on that line. For multiple points (or continuous paths), it's trickier, and I'm not sure how to say it other than "express the points as unit vectors and take the average coordinate wise".

Cheers, Wayne
 
jaggedben said:
Okay, great, but what's the answer?
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Hey, I'm a mathematician, that means the problem is solved as soon as I figure out how to calculate the answer, there's no need to actually do the calculation. : - )

More seriously, I'm working it out symbolically based on the equations for the path of the sun in the sky, it may take me a little while.

Cheers, Wayne
 
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