B Why can't Solar panels be hemispherical, or a curved strip type?

[russ_watters]

Why is there lower power in the middle of the day? I thought they oriented at the optimal angle at all times. How can you do better than that given a lack of space or desire for a larger panel?

You lost me when you said half a hemisphere. Did you mean half a sphere? Because half a hemisphere has the same area as a circle of the same radius.

Single axis means they trace one arc across the sky. The sun traces a different arc every day. I'd presume that the optimal arc for maximizing annual energy production is somewhere between maximum summer and maximum winter peak power.

 

[mongotuslan]

Why is there lower power in the middle of the day? I thought they oriented at the optimal angle at all times. How can you do better than that given a lack of space or desire for a larger panel?



You lost me when you said half a hemisphere. Did you mean half a sphere? Because half a hemisphere has the same area as a circle of the same radius.

The lower power in the middle of the day is due to the single axis oriented horizontally, North to South, so there is Cosine effect due to the Sun angle at Noon from the Latitude. At some time during the morning and afternoon, the single axis tracking will align the panel normal to the Sun.

I integrated form 0 to 90 degrees, half a hemisphere, instead of -90 to +90 degrees, a full hemisphere, and assumed symmetry from this arc to a hemisphere. I am not sure I did this correctly, which is why I am asking for someone to check my math.

 

[renormalize]

I integrated form 0 to 90 degrees, half a hemisphere, instead of -90 to +90 degrees, a full hemisphere, and assumed symmetry from this arc to a hemisphere. I am not sure I did this correctly, which is why I am asking for someone to check my math.

I don't see enough equations in any of your 3 posts in this thread for readers to check. Don't make us guess! Learn enough LaTeX to post your work here, including basic assumptions, equations, and the form of the integral you evaluated numerically by spreadsheet. A diagram of the geometry you're considering would also be helpful.

 

[Herman Trivilino]

The lower power in the middle of the day is due to the single axis oriented horizontally, North to South, so there is Cosine effect due to the Sun angle at Noon from the Latitude. At some time during the morning and afternoon, the single axis tracking will align the panel normal to the Sun.

Okay, I didn't pay attention to what "single axis" means. Got it now. I was under the mistaken impression that the tracking system compensated for the seasons.

I integrated form 0 to 90 degrees, half a hemisphere, instead of -90 to +90 degrees, a full hemisphere, and assumed symmetry from this arc to a hemisphere. I am not sure I did this correctly, which is why I am asking for someone to check my math.

Well, I did as close to that as I could. If you were thinking that half a hemisphere has twice the area of a circle of the same radius, that would have been an error.

As @renormalize points out, we can't check your math because you didn't show us your math. You described the math you did, but describing what you did is not the same thing as showing what you did.

 

[256bits]

I drafted a numerical integration in a spreadsheet over half a hemisphere, and came up with a factor of 0.635. That is, the effective illuminated area of a hemisphere is 0.635 the geometric surface area

The 0.635 seems awfully close to the average value of a half cycle of a sign wave, from 0 to pi, which should be the same for a quarter cycle from 0 to pi/2.

also, please define 'effective illuminated area'. I am not sure what is meant.

 

[sophiecentaur]

I’m assuming that the driving force in choice of panel shape has been large areas of panels.
It may have been mentioned above but I’ll say it anyway. At low incidence, the mutual shading would reduce output of many 3D panels. Offsetting could reduce this but when the Sun is low, the performance would be compromised at different times of year.

Cylindrical arrays are great for some rotating satellites but, otherwise they don’t get my vote.

 

[Ken Fabian]

Now it is more cost effective to over-build and just add more fixed PV. Flat panels are easiest to make and to install. It is almost exclusively 'fixed' flat panels now; tracking systems are rarely cost effective.

Lack of space/shading is a site specific problem best fixed by choosing a different site - not a problem for large scale ground mounted solar farms but frustrating for homeowners unable to take advantage. But if they are connected to the grid then daytime electricity here (Eastern Australia) will have a large proportion of solar, from rooftop feed-in as well as from ever more solar farms.

We do see some 'afternoon' West oriented PV, usually one 'string' (many solar inverter-controllers handle 2 strings, to make upgrading easier?), usually in combination with a string North oriented (note, Southern Hemisphere here) but I suspect it is done more because the roof hasn''t the room for all facing North than to maximise afternoon output closer to the evening peak demand.

Batteries are the emerging means to maximise effective usage beyond sunlight hours. (Just saw that Aldi is selling solar/battery packages for Australian homes, a lot cheaper than ours was, sigh - yet we have done okay cost wise)

We upgraded to larger battery recently and effectively run our home entirely off solar - whilst still sending about 2x what we use to the grid. Enough battery for all but exceptional circumstances, but the solar inverter is a bit undersized. I expect with an EV with vehicle to home connection we would have 365 day/night a year electricity.

 

[Baluncore]

On pitched roofs, we have a constraint that requires the panels not protrude significantly beyond the existing roof profile. That reduces both the visibility and the profile that might be under-cut, and so lifted in high winds. Any panel shape, other than flat, would exceed those constraints. It would also require a wind engineering analysis.

 

[mongotuslan]

Okay, I didn't pay attention to what "single axis" means. Got it now. I was under the mistaken impression that the tracking system compensated for the seasons.



Well, I did as close to that as I could. If you were thinking that half a hemisphere has twice the area of a circle of the same radius, that would have been an error.

As @renormalize points out, we can't check your math because you didn't show us your math. You described the math you did, but describing what you did is not the same thing as showing what you did.

I completely redid my numerical integration spreadsheet, and the results seem to support that the Cosine Effect on a hemispherical PV array would have the same output as a disc with the same radius normal to the light rays from the Sun. I made a prototype spreadsheet with the formulas for two adjacent latitude lines. The diagram is at this link:
https://en.wikipedia.org/wiki/Spherical_coordinate_system
Spherical Latitudinal Strip Solar Array with Cosine Effect.
Rays of sunlight are parallel to the polar z-axis.

The spreadsheet has 91 rows, with the first column the THETA degrees from 0 to 90 in 1 degree increments. Each row calculates the area of a strip of latitude 1 degree wide. The sum of the rows for r=1 is extremely close to pi, and given the granularity error of numerical integration, this supports the hypothesis that the Cosine effect area of a hemisphere is the same as the area of the equatorial disc.
For a spacecraft, a spherical array of PV panels will provide the same solar power as a disc of the same radius without a Sun tracking mechanism, but at a cost of 4 times the number of PV panels.