Comment by fbdab103
1 year ago
Wondering if anyone could help shore up my understanding or point out a resource to get me a better handle on this.
Using the solar maps from here[0], you can find the kWh/day/m2 for the US. If I am in a say 5.7 kWh/day/m2 region and I have 1 m2 of a 20% solar efficiency panel, does that mean I would get 1.14 kWh usable out the other end? Or is it 20% * X% horribly lossy conversion factor?
If I want to math out 11kWh/day in the 5.7 region, back calculating would put me at requiring 9.6 m2 of panels (11 / (5.7 * 0.2). Again, if there is a horrible lossy conversion factor, that would just go into my denominator, correct?
Or am I missing something entirely? I tried to use this calculator[1], but I could not recapitulate the numbers they were generating.
You'd be able to generate 1.14 kWh at the panel level if you kept the panel pointed directly at the sun throughout the day [1]. This is called "2 axis tracking" and it was sometimes used for solar farms when solar panels were much more expensive. Now that panels are much cheaper, 2 axis tracking has practically vanished from the market. The added expense and mechanical complexity isn't worth it. Single axis tracking, where the panels just rotate to track the sun from east to west, is still popular in large solar farms. It captures more sun than leaving the panels stationary but has less complexity than 2 axis tracking.
For a rooftop solar panel, you're not going to have any sort of sun tracking. The lack of tracking will reduce your output at the panel level. You will also lose more output if dust, debris, and bird droppings don't get cleaned away regularly.
You also lose some energy when the direct current electricity from your panels gets converted to alternating current in the inverter. How much loss depends on the inverter and how heavily loaded it is.
The NREL tool you linked says it's designed for "homeowners, small building owners, installers and manufacturers", which implies that it's for rooftop systems. It includes estimates for those loss factors I mentioned above, which is why I expect that it falls short of the number you calculated.
[1] EDIT: I forgot another significant factor: temperature coefficient of performance. A panel gets its efficiency measured at "standard test conditions" which include a moderate (near room temperature) panel temperature. Panels lose some efficiency as they heat up, which means that they don't perform as well as you might naively expect in the middle of the summer. The loss varies by panel technology. The very best conditions for panel output -- where they actually surpass reported efficiency -- is "bright sun but cold air," like noon on a freezing cold day with clear skies.
So, the bottom line is that the simple kWh/day/m2 * panel efficiency * m2 of panels should be within the theoretical ballpark of generation, but the real world is a harsh mistress and will undercut you.
On that calculator resource, they provide a monthly and hourly spreadsheet, but even with the more detailed numbers, I was still failing to corroborate their presumably much more sophisticated modelling which accounts for other losses.
Thanks. Just spit balling numbers and trying to see what things look like.
> which include a moderate (near room temperature) panel temperature. Panels lose some efficiency as they heat up, which means that they don't perform as well as you might naively expect in the middle of the summer.
This is why vertical solar panels are becoming a thing, the additional cooling benefits increase output up to or beyond the optimal angle to the sun, and the better cooling also prolongs their life.
And they work better in winter if there's snow.
The "System Losses" breakdown shows the various additional factors they are derating the system by. The figures seem reasonable enough, and give an additional loss of 14%.
I put in my own address, which is in the 4.0-4.5 kWh/day region, and set the DC system size to 1 kW, which corresponds to 6m2 of panels (courtesy of their rooftop calculator). The NERL website estimated that such a system would yield between 2.45 and 6.48 kWh/day, with an annual mean of 4.71 kWh/day.
That works out pretty close to what the map indicates for my region: 4.5 kWh/m2day * 0.2 conversion factor * 0.86 losses factor * 6 m2 of panels = 4.64 kWh/day
Super appreciate your reverse calculation. The downstream 14% loss still is an easy enough fudge factor to use for a plausible output number.
The map shows solar resource, irrespective of the technology you're using. For example, you could be using heat collectors that would capture closer to 100%. To keep things simple, for back of the envelope calculations, you can imagine 1kW per square meter. Subtract cloud coverage, night hours and then multiply with 0.2 for PV panels efficiency.
If you want 11kWh/day, you need: (5.7 * 0.2) * Y = 11, so Y = 10 square meters. You can double check this: 10 sqm should have about 10KW of solar potential energy, but with PV efficiency you're getting about 2KW, so to reach 11kWh, you need 5 good hours of sunshine on average.