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What is the formula to caclulate or estimate the electricity generated in output of a solar panel system?
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Google "PVWatts", read the help screens/documentation, and come on back here with your results or any questions.CS6P-260P/SE3000 - http://tiny.cc/ed5ozx -
Couple things to assume (starting with Grid tie batteryless systems and then to battery/off-grid systems)
1) PV panels produce (generally) 80% of their STC wattage
2) Grid Tie inverters are 95% efficient
3) Battery charge controllers are
MPPT 95% efficient when in BULK mode only, otherwise they are -
PWM efficiency varies with the the ratio of PV array output voltage and battery charge voltage
4) Flooded batteries start life at 90% efficient, decreasing to 70% at failure
5) Off Grid inverter efficiency varys from 70-95% depending on loading and brand/model
Hope this quick guess works for you
Powerfab top of pole PV mount (2) | Listeroid 6/1 w/st5 gen head | XW6048 inverter/chgr | Iota 48V/15A charger | Morningstar 60A MPPT | 48V, 800A NiFe Battery (in series)| 15, Evergreen 205w "12V" PV array on pole | Midnight ePanel | Grundfos 10 SO5-9 with 3 wire Franklin Electric motor (1/2hp 240V 1ph ) on a timer for 3 hr noontime run - Runs off PV ||
|| Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||
|| VEC1093 12V Charger | Maha C401 aa/aaa Charger | SureSine | Sunsaver MPPT 15A
solar: http://tinyurl.com/LMR-Solar
gen: http://tinyurl.com/LMR-ListerComment
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Hi Mike90250,
I want to make sure I understand your information correctly - you say PV panels generally produce about 80% of their STC (standard test conditions) wattage. So if I had a 300w panel (module) I should actually expect just 240w in real world conditions, (not really 300w as I might get under optimal conditions in a lab test).
Then, I should expect to convert only 95% of the panel (module) output from DC power to AC power. This is just because of the way electrical engineering and physics works (?).
So in summary, at any moment in time, one 300w panel would ordinarily be expected to generate 240w (as DC power?) ... and that 240w would then be converted to AC by the inverter (losing 5% in the process) resulting in something like 228w of AC power? So it is really just the 228w times hours of production that I am calculating to figure out my offset?
Thanks.Comment
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No... that 80% number is more like the peak output you will get near the middle of the day, if you have south facing panels with a tilt about equal to your latitude. Once you get more than an hour or so away from solar noon, that number starts to fall off fairly quickly. In the spring when it is cooler, the peak might be a bit higher into the mid-80's, in the winter, maybe not even 70%, again, depending on the orientation.
PVWatts is a great tool to estimate how much array is needed to achieve a targeted offset, and the hourly output option can be used for more granular analysis if needed.CS6P-260P/SE3000 - http://tiny.cc/ed5ozxComment
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Well... except that the sun varies with time, too.
Take a look at http://pvwatts.nrel.gov, that implements the formula you're looking for, and takes the sun into account (assuming no shading).
(Watch out though - it gives the output for a typical year with both cloudy and sunny days, so don't take any one day's output too literally.)
And/or look at a nearby similar system, find out how many kWh it generated last year, divide that by the sum of the faceplate ratings of its panels, and you've got the infamous kWh/kW aka "sun hours" aka "yield" number (see eq 1 in http://www.nrel.gov/docs/fy05osti/37358.pdf ), also called 'efficiency' by pvwatts.org ( see my system's at http://pvoutput.org/aggregate.jsp?id...=44908&v=0&t=y )
This is roughly the "hours of production" you're talking about, but it already includes the inefficiencies of that system's inverter.
For my system over a whole year, it's about 4.5, but your mileage will vary.
Comment
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Ok, so it sounds like a general rule of thumb is that any given panel will just give you up to 80% of the rated power at the peak moment in the day... And ordinarily a little less depending on conditions and time of day.
And that from that number (being generated at the panel) you will lose 3-5% at the inverter.
So you may build a 7.68Kw system... But don't be surprised if under optimal conditions,the readout from the system just says it is making 6200w AC.
?Comment
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My system's nameplate output is 8950 watts DC. Today, a hot sunny day, the AC output was
8am 1000W
9am 2300W
10am 3800W
11am 4900W
noon 5700W
1pm 6100W
2pm 6100W
3pm 5700W
...
Peak today was about 68%, not 80%.
At no point has it ever gotten anywhere close to 8900W AC. Best I've ever seen it do is 7800W, but that's rare.
See graph at http://pvoutput.org/intraday.jsp?id=...08&dt=20160726Comment
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Besides time of day, clouds might knock your output down to as low as 10% of normal. Between clouds and shorter days, my
winter months may produce only 1/3 as many KWH as summer months. Bruce RoeComment
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No Dan it isn't. That is an incorrect, misleading and ignorant statement.
Sun hours" is generally considered to be the number of kWh/m^2 of solar radiation falling on a horizontal surface per time period (usually per day).
The term "sun hours" has nothing to do with system performance or system size. It is an anachronistic term that represents resource availability only. The currently accepted term is Global Horizontal Radiation or GHI.
Get it right.Comment
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Thanks for the info everyone.Comment
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Powerfab top of pole PV mount (2) | Listeroid 6/1 w/st5 gen head | XW6048 inverter/chgr | Iota 48V/15A charger | Morningstar 60A MPPT | 48V, 800A NiFe Battery (in series)| 15, Evergreen 205w "12V" PV array on pole | Midnight ePanel | Grundfos 10 SO5-9 with 3 wire Franklin Electric motor (1/2hp 240V 1ph ) on a timer for 3 hr noontime run - Runs off PV ||
|| Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||
|| VEC1093 12V Charger | Maha C401 aa/aaa Charger | SureSine | Sunsaver MPPT 15A
solar: http://tinyurl.com/LMR-Solar
gen: http://tinyurl.com/LMR-ListerComment
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