Should I insist on a second inverter?

[Shiro]

I'm currently doing the design for my system. I'll be going with a 12kw system with 2 powerwalls. All of my panels will face south or east, there's nothing shading the roof, and this is sunny SoCal so sunlight isn't an issue.

Should I insist on a second inverter - maybe a 7.6 and a 3.8? Or will the production increase not outweigh the cost of the inverter?

 

[wwhitney]

What's the roof slope, and what's the division of panels between east and south, and is south due south?

Cheers, Wayne

 

[Shiro]

30 panels, one array of 7 facing approx due south and an array of 23 facing approx 136 degrees SE. don’t know the roof pitch but it’s a 60’s ranch style house.

edit: sorry, not due south. It’s about 211 degrees SE for the 7 panel array.

 

[wwhitney]

Well, the simple geometric analysis depends on knowing the azimuths and roof pitches fairly exactly. I'm going to assume you mean 0 degrees azimuth = due north. So 211 degrees is SSW, and 136 degrees (a 75 degree difference, not 90?) is SE.

If your roof pitch is 4:12 (a total guess), then that's an angle of arctan(4/12) = 18.4 degrees. You can represent each array by a vector, whose direction is normal to the array and whose magnitude is the number of panels (if they are all identical).

Then array 1 is 7 * (sin 18.4 * cos (90 - 211), sin 18.4 * sin (90 - 211), cos 18.4) = ( -1.14, -1.89, 6.64). And array 2 is 23 * (sin 18.4 * cos (90- 136), sin 18.4 * sin (90 - 136), cos 18.4) = (5.04, -5.22, 21.82). The bit about "90 - azimuth" changes from the compass convention to the math convention where E = 0 degrees, N = 90 degrees, etc.

So when both arrays are illuminated, they will jointly act like an array represented by the sum of those vectors, or (3.90, -7.11, 28.46). That vector represents an array of 29.6 panels (the norm of the vector, sqrt(x^2 + y^2 + z^2) ) at an elevation angle of arccos(28.46/29.6) = 15.9 degrees and an azimuth of 90 - arctan(-7.11/3.9) = 151 degrees

Now you plug your location into PVWatts and model an array of size 12 kW * (29.6/30) with those parameters, with the different DC/AC ratios from a single 7.6 kW inverter, or a 11.4 kW of inverter, to see what the difference is. Except you need to confirm the azimuths and roof pitch, and adjust my computation accordingly.

Cheers, Wayne

[I went through the computations because I haven't actually done it before. I'm slightly surprised that for a roof pitch of 4:12 and an azimuth difference of 75 degrees, the equivalent array is still of size 29.6 panels. I did double check the computations, but there could still be an error.]

 

[Shiro]

I plugged the numbers into PVwatts, and if I've done this right, I lose about 940 kW/year due to the inverter size, so somewhere between 4 and 5%?

Do those numbers sound like a second inverter is worth it in the long run?

 

[holeydonut]

I plugged the numbers into PVwatts, and if I've done this right, I lose about 940 kW/year due to the inverter size, so somewhere between 4 and 5%?

Do those numbers sound like a second inverter is worth it in the long run?

Yeah, I don't understand Wayne's math (he lost me the moment he used trigonometry since my brain cells for math died many decades ago). I think going to PV Watts and changing the advanced stats from a 1.57 DC:AC ratio to 1.05 DC:AC is easier heh.

940 kWh per year x 20 years (the approximate useful life of your system) x $0.25 per kWh (rough estimate to what your electricity costs are) would be $4,700. Granted, this means you plan on being in your house for the long haul. If you're going to sell or move out of your home in a few years don't sweat it.

I can't imagine Tesla charging you more than $4,700 for a 3.8 kW inverter. Although your bigger hurdle is that Tesla probably won't want to make the change because their AI Robots would crash into traffic cones or the back of a parked Sheriff's cruiser while processing your change order.

Are you sure you can't fit a few more panels on your roof? Solar is so cheap now even some north facing panels in SoCal + the 3.8 kW inverter could make your system sized for future EVs and make lots of economic sense.

 

[Shiro]

Yeah, I don't understand Wayne's math (he lost me the moment he used trigonometry since my brain cells for math died many decades ago). I think going to PV Watts and changing the advanced stats from a 1.57 DC:AC ratio to 1.05 DC:AC is easier heh.

940 kWh per year x 20 years (the approximate useful life of your system) x $0.25 per kWh (rough estimate to what your electricity costs are) would be $4,700. Granted, this means you plan on being in your house for the long haul. If you're going to sell or move out of your home in a few years don't sweat it.

I can't imagine Tesla charging you more than $4,700 for a 3.8 kW inverter. Although your bigger hurdle is that Tesla probably won't want to make the change because their AI Robots would crash into traffic cones or the back of a parked Sheriff's cruiser while processing your change order.

Are you sure you can't fit a few more panels on your roof? Solar is so cheap now even some north facing panels in SoCal + the 3.8 kW inverter could make your system sized for future EVs and make lots of economic sense.

I could fit more panels for sure but this setup even with the single inverter projects to meet 111% of my energy requirements and I'll also be getting 2 powerwalls.

Not sure if I'll be in this house for 20 years...it's a possibility but not a certainty.

Maybe the single inverter is the way to go.

 

[jhn_]

this thread

Math and logic for the win!

 

[Redhill_qik]

I plugged the numbers into PVwatts, and if I've done this right, I lose about 940 kW/year due to the inverter size, so somewhere between 4 and 5%?

Do those numbers sound like a second inverter is worth it in the long run?

It depends. If you are a net consumer than that would be worth 940 kWh * $0.40/kWh or $376/year and should pay for itself in 5 years. If you are a net generator then that would be worth 940 kW * $0.045/kWh or $42.30/year and probably wouldn't be worth it over 30 years, but might be worth it as a redundancy feature so you could still be operating in one of the invertors died.

 

[wwhitney]

I plugged the numbers into PVwatts

The numbers you got are probably a worst case in terms of clipping, as your roof slope may be greater than 4:12, and I bet the azimuth difference is actually 90 degrees, rather than 75 degrees.

A worst case for production would be if the roof planes are actually 90 degrees apart in azimuth (as expected), and the elevation angle is 90 degrees (i.e. they are walls, not roofs). Then your two arrays are pointing in perpendicular directions, so the equivalent array size is juts sqrt(23^2 + 7^2) = 24 panels. So the nominal DC/AC ratio wouldn't be 12 kW / 7.6 kW = 1.58 but 80% of that, or 1.26. Plus with the low angle (0 degrees) of the normal, the sun would be quite a bit off axis at all times, so you'd get no clipping.

That's obviously not what you have, but it shows the maximum possible extent of the effect (assuming the azimuth angle is not more than 90 degrees). Let me do one more example, with numbers simpler to computer with. Say you 7 panels SW (azimuth 225) and 23 panel SE (azimuth 135) and the roof pitch is 6.9 : 12 (30 degree tilt). Then the unit normals are (+/- sin 30 * cos 45, - sin 30 * sin 45, cos 30) = (+/- sqrt(2)/4, sqrt(2) /4, sqrt(3)/2). And the sum of the arrays would be ((23-7) sqrt(2)/4, -30 sqrt(2)/4, 30 sqrt(3)/2). With the equivalent array size = sqrt( 16^2 / 8 + 30^2 / 8 + 30^2 * 3/4) = 28.6

Cheers, Wayne

 

[RichardCullip]

Wayne has thrown some serious math at us and come up with a good approximation to use in PVWatts. An easier, but not necessarily more accurate, method would be to perform two PVWatts runs, one for each cluster of panels, and add the results together.

 

[patrick40363]

I would ask for the second inverter.

 

[wwhitney]

An easier, but not necessarily more accurate, method would be to perform two PVWatts runs, one for each cluster of panels, and add the results together.

A less easy but more accurate (than my geometric only approximation, as it takes into account weather estimates) method is to perform two PVWatts runs, dump the full 8760 hourly data points, and then add those pointwise. Then you can compare that result with the version where you also cap the AC output at the inverter rating, and see what the difference is.

Cheers, Wayne

 

[holeydonut]

A less easy but more accurate (than my geometric only approximation, as it takes into account weather estimates) method is to perform two PVWatts runs, dump the full 8760 hourly data points, and then add those pointwise. Then you can compare that result with the version where you also cap the AC output at the inverter rating, and see what the difference is.

Cheers, Wayne

You can also cross-check things with the AESC calculator that the IOUs use to administrate the SGIP.

CSI EPBB Calculator - Standard PV Calculator

 

[h2ofun]

Wayne has thrown some serious math at us and come up with a good approximation to use in PVWatts. An easier, but not necessarily more accurate, method would be to perform two PVWatts runs, one for each cluster of panels, and add the results together.

I did this for my panels, which are connected as 6 sets on 2 inverters. I did 6 calculations and added together

 

[Shiro]

Just checking back in with an update.

I've settled on a final design and managed to get them to write up my system as two Powerwall +'s (which each come with an inverter) and my estimated annual production number jumped up over 2k kWh/year. System cost did not change so I'm very happy with this outcome.

Now begins the waiting game for permit, install, pto.