Confusion re module rating, microinverter rated output power, and PVWatts.

[NEOH]

You didn't run the model with the same inputs as the op, so you didn't get the same outputs. The conditions you ran are showing clipping. The conditions the OP ran didn't show clipping (high loss factor + low tilt for the latitude), so the difference in inverter loading affected the OP's output. RTFM if you want to see the equations involved, but both your output and the OP's are "correct" for the inputs used.

OK,
I re-ran the PV Watts for 20% Loss instead of 10% losses
The change from 10% Losses to 20% Losses does change the output from PV Watts.
So, increasing the "Losses" field essentially lowers the "NET" DC-to-AC ratio used in the calculations vs what we actually entered

[sensij]

I don't know?
Because I agree with you.
The OP's numbers do not agree with my PV Watts numbers.

You didn't run the model with the same inputs as the op, so you didn't get the same outputs. The conditions you ran are showing clipping. The conditions the OP ran didn't show clipping (high loss factor + low tilt for the latitude), so the difference in inverter loading affected the OP's output. RTFM if you want to see the equations involved, but both your output and the OP's are "correct" for the inputs used.

Edit : if you continue to drop the ratio further (0.9, 0.8, etc), you should eventually see the energy start to drop, as the OP did.

[NEOH]

Now I am confused.
How is it that the output is getting LOWER as the DC to AC ratio approaches 1?

I don't know?
Because I agree with you.
The OP's numbers do not agree with my PV Watts numbers.

From PV Watts ...

Production Ratio
========== =====
5,093 KWHr 1.1 <<< 4KW PV with 3.6 KW Inverter (Per PV Watts help file)
5,092 KWHr 1.2 <<< 4KW PV with 3.3 KW Inverter
5,073 KWHr 1.3 <<< 4KW PV with 3.1 KW Inverter
5,031 KWHr 1.4 <<< 4KW PV with 2.9 KW Inverter = Smallest inverter

It appears that PV Watts assumes a constant PV DC Size, while the Inverter AC size gets smaller as the DC-to-AC Ratio increases from 1.1 to 1.4.
PV Watts shows me lower and lower production with increased DC-to-AC Ratios, ie smaller and smaller Inverter.
Smaller Inverter = Lower Production = Very logical.

[ButchDeal]

Thanks, as always. I say 'bah' to the inverters for the moment at least. That is one of the things I really don't embrace about SAM. In PVWatts, I can just say give me a 1kW array, in Sam I have to go specify 4 Canadian Solar CS6J-250LMNOPs (or find something else to represent 1 kW). In PVWatts, I can just say give me a decent default inverter, in SAM I have to go pick out the specific inverter(s) from a list of hundreds and just pray they represent something that SAM won't balk on.

Don't get me wrong, I understand and admire the power in being able to simulate real world more closely, but if I am only interested in some of the DC aspects of a situation, the lack of a 'unspecified' panel or 'blatantly average' inverter is a pain. At least SAM will let me proceed with 'No Financial Model', if it didn't I would probably chuck it 90% of the time.

It is simulating real world "more closely" it is actually simulating real world, where pvwatts is only doing a half baked calculation based on rule of thumb.
making advanced calculations and charts based on simple rule of thumb, and not understanding the limitations of such a rule gets people to make silly statements.

For a more advanced real world simulation we use Aurora which does all of the electrical as well as 3D shadow rendering calculations for a true real world simulation.

[AzRoute66]

I think it is because of the way PVWatts handles inverter efficiency. [...] As usual, if you want a more sophisticated (inverter specific) model, you have to step up to SAM.

Thanks, as always. I say 'bah' to the inverters for the moment at least. That is one of the things I really don't embrace about SAM. In PVWatts, I can just say give me a 1kW array, in Sam I have to go specify 4 Canadian Solar CS6J-250LMNOPs (or find something else to represent 1 kW). In PVWatts, I can just say give me a decent default inverter, in SAM I have to go pick out the specific inverter(s) from a list of hundreds and just pray they represent something that SAM won't balk on.

Don't get me wrong, I understand and admire the power in being able to simulate real world more closely, but if I am only interested in some of the DC aspects of a situation, the lack of a 'unspecified' panel or 'blatantly average' inverter is a pain. At least SAM will let me proceed with 'No Financial Model', if it didn't I would probably chuck it 90% of the time.

[sensij]

Now I am confused.

How is it that the output is getting LOWER as the DC to AC ratio approaches 1? I thought 1 was the epitome of 'no clipping'.

I think it is because of the way PVWatts handles inverter efficiency. A higher DC to AC ratio implies a smaller inverter, and a smaller inverter running closer to 100% will be modeled as more efficient than a larger inverter that is not so heavily loaded. The output shown is basically saying there is no difference in clipping, so the inverter efficiency is driving the change in output. The differences are so small that I wouldn't assign much weight to them. As usual, if you want a more sophisticated (inverter specific) model, you have to step up to SAM.

[AzRoute66]

[...]

Annual output 6,604 kwh @ 1.35 ratio
Annual output 6,603 kwh @ 1.3 ratiio
Annual output 6,600 Kwh @ 1.25 ratio
Annual output 6595 kwh @1.16 ratio (this would be the ratio with the m250 microinverter paired with 290W panels)

Now I am confused.

How is it that the output is getting LOWER as the DC to AC ratio approaches 1? I thought 1 was the epitome of 'no clipping'.

[sensij]

Again, please correct me where I am misguided.

Based on what you've written, I think your M215 vs M250 decision making is fine. I'm curious about the comparison of M215 costs vs a string inverter or Solaredge system.

Have you budgeted in the trunk cable? With a max of 17 M215's per circuit, will you be running two circuits for the SSW array, and one for the ESE array, or can you wire the arrays together for just two circuits? Does your AHJ expect to see a separate subpanel to combine those circuits?

[wexcellent]

I just ran the numbers again using your 8% loss instead of my 18%. In that scenario with all other parameters identical to above the M250 would produce slightly more power than the M215 (23 kwh annually for my system). However, using the M250 over the M215 would cost me an additional $1200 or so. I don't have much of a grasp about how loss is calculated. Perhaps 8% is idealistic and 18% is pessimistic? PVWatts default is 14% but I guess microinverters mitigate some of that? Anyway, even best case scenario of 8% loss, 23 kwh a year isn't worth the $1200, especially if my loss is actually greater and thus the M215 outperforms the M250 in my installation. Again, please correct me where I am misguided.

[sensij]

18% and 13% loss? That is really high. Most properly designed systems are modeled well by 10% or slightly less.

M215 produces 225 W continuously.

I don't disagree with the conclusion you've made... For the right price, M215's are a reasonable choice. Just think through the long term implications of 31 eventual inverter failures on your roof, with no inverters still in production that are compatible out of the box with the interconnect the M series uses.

[wexcellent]

So I read the relevant parts of "Solar Power Your Homes For Dummies" and found little that helped me understand how clipping works.

Because of cost and reliability I plan to stay with the Enphase M215 so I ran different scenarios with PVWatts at my location. I will have two different rooftop arrays, location is Troutdale, OR. Here are my results:

SSW Array, minor spring/fall shading, 18% loss, standard module, roof mount
(21) 290W rated panels = 6.09 kw, 96.5% efficient M215 microinverters rated for 215W continuous results in a DC/AC ratio 1.35
22.62 tilt, 193 Azimuth
Annual output 6587 kwh @ 1.5 ratio
Annual output 6603 kwh @ 1.4 ratio
Annual output 6,604 kwh @ 1.35 ratio
Annual output 6,603 kwh @ 1.3 ratiio
Annual output 6,600 Kwh @ 1.25 ratio
Annual output 6595 kwh @1.16 ratio (this would be the ratio with the m250 microinverter paired with 290W panels)

ESE Array, no shading, 13% loss, standard module, roof mount
(10) 290W rated panels = 2.9 kw, 96.5% efficient M215 microinverters rated for 215W continuous results in a DC/AC ratio 1.35
22.62 tilt, 103 Azimuth
Annual output 2,917 kwh @ 1.4 ratio
Annual output 2.918 kwh @ 1.35 ratio
Annual output 2,917 kwh @ 1.3 ratiio

Like I said, I don't really understand how this whole clipping thing works but according to this it seems like the sweet spot for me is indeed pairing an M215 with a 290W panel, as the bid I obtained suggested. If I pair a M250 with a 290W panel my DC/AC ratio would be 1.16 and my annual output would be less.

Please help me understand where I am misguided.

[J.P.M.]

While I appreciate the point you'd like to make, I'm having trouble replicating your results.

Max continuous output of an M215 = 225 W. With a 290 W panel, that is a DC to AC ratio = 1.29.
Max continuous output of an M250 = 250 W. With a 290 W panel, that is a DC to AC ratio = 1.16.

PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.00 ratio = 422 kWh / yr
PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.16 ratio = 421 kWh / yr (0.25% loss to clipping)
PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.29 ratio = 417 kWh / yr (1.2% loss to clipping)

Have I made different assumptions in the model than you did?

Later.

I made some errors with respect to clipping losses. Thank you for catching them.

My apologies to Inspron and Wexcellent for any confusion and waste my errors may have caused.

When I calc'd an annual clipping estimate I measure against a system (0.290 W in this case) with an assumed string inverter using PVWatts default DC to AC size ratio of 1.0. Then, for any hourly output that's > the micro equipped max output, I subtract the lesser (225 Watts) micro output from the string inverter equipped output. The sum of the 8,760 hourly diff. calculated in that fashion is the annual clipping estimate. That's perhaps the Luddite way, and I do understand DC to AC ratios as PVWatts works, but this way allows an easy read of clipping hours and it's just easier to keep straight in my head.

My error was reading the wrong column in my calculations - the one with the hourly total before the micro max. capacity subtracted, and partly, but certainly not entirely a consequence of not having someone to check my work as was a requirement when I was gainfully employed.

Redoing the numbers, my estimate, using PVWatts, 10 % system losses, 180 deg. az., 45 deg. tilt, 417 kWh/yr. output is that the M215's will clip ~ 210 - 220 hours/yr. and gather ~ 4 kWh less per 290 Watt panel than a string inverter equipped system. The M250's will clip ~ 65-70 hrs./yr. and overall produce about 1 kWh less over a year, or essentially no clipping.

Those estimates seem to agree with yours.

I do still maintain however that an array's annual performance will be mostly unaffected solely by how far it is from the equator. Off grid performance will be affected by seasonal irradiance variation as it may affect battery performance (or lack of it) brought on as f(latitude) and the shorter days that go with higher latitudes and how short days/long nites affect batteries, but grid tie systems in sunny northerly climates (like the Yakama valley in WA, or northern NV for example) will, probably outproduce similar systems on an annual basis in, for example, the southeastern U.S.

Provided each array is optimally oriented for the location it's in, a higher local solar clearness index and lower temps. will make the difference more than latitude.

[J.P.M.]

While I appreciate the point you'd like to make, I'm having trouble replicating your results.

Max continuous output of an M215 = 225 W. With a 290 W panel, that is a DC to AC ratio = 1.29.
Max continuous output of an M250 = 250 W. With a 290 W panel, that is a DC to AC ratio = 1.16.

PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.00 ratio = 422 kWh / yr
PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.16 ratio = 421 kWh / yr (0.25% loss to clipping)
PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.29 ratio = 417 kWh / yr (1.2% loss to clipping)

Have I made different assumptions in the model than you did?

I might have screwed up. More later.

[sensij]

For arrays with close to an optimal orientation with respect to tilt , latitude has little to do with it until the location gets ~ > 60 N. or S. latitude.
...

Unless I had a lot of shade, I'd go with a string inverter.

While I appreciate the point you'd like to make, I'm having trouble replicating your results.

Max continuous output of an M215 = 225 W. With a 290 W panel, that is a DC to AC ratio = 1.29.
Max continuous output of an M250 = 250 W. With a 290 W panel, that is a DC to AC ratio = 1.16.

PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.00 ratio = 422 kWh / yr
PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.16 ratio = 421 kWh / yr (0.25% loss to clipping)
PVWatts for Portland, ME TMY3, 180 deg Az, 45 deg tilt, roofmount, premium panels, 8% loss, 1.29 ratio = 417 kWh / yr (1.2% loss to clipping)

Have I made different assumptions in the model than you did?

[J.P.M.]

If you are very far north, it is possible that M215 could work since you are unlikely to achieve outputs in excess of 215W even in the summer with the 290W panels.

For arrays with close to an optimal orientation with respect to tilt , latitude has little to do with it until the location gets ~ > 60 N. or S. latitude.

One example: Portland ME:

PVWatts

M215 on a 290 W panel will clip 330 hrs. of 4510/yr., 78 kWh of 417 kWh/yr., or ~ 18% loss of total possible output.

M250: on a 290 W panel will clip 66 hrs. of 4510 hrs., 17 kWh of 417 kWh/yr., or ~ 4 % loss of total possible output.

Azimuth 180, tilt 45 deg.

Unless I had a lot of shade, I'd go with a string inverter.