AC system losses

[nerdralph]

Today I made a rough measurement of AC system losses from my inverter to the grid. I recorded the voltage and current at the inverter during bright sun shortly after solar noon, switched off the DC input, waited for a few seconds while the output current dropped to 0, then recorded the voltage again. In bright sun it was 252V/17A, and 245V/0A, which works out to .412 Ohms and 2.8% system losses at peak output.
Has anyone done similar measurements?

[J.P.M.]

Closest I have is from the inverter screen as one of a bunch of variables I measure as part of my array fouling measurements.

Those data, about 16 or so array, inverter and environmental variables, including GHI and conversion algorithms for POA irradiance, include the input from the array and the output of the inverter.

Typical inverter output to input ratios of power under cloudless conditions at the minute of the daily time minimum angle of incidence of beam irradiance on my array typically runs about 0.970 to 0.971, mostly independent of everything I'm measuring at that time. I've done that more than 500 times or so over the last 8+ years and the inverter efficiency hasn't been outside those limits.

[foggysail]

Today I made a rough measurement of AC system losses from my inverter to the grid. I recorded the voltage and current at the inverter during bright sun shortly after solar noon, switched off the DC input, waited for a few seconds while the output current dropped to 0, then recorded the voltage again. In bright sun it was 252V/17A, and 245V/0A, which works out to .412 Ohms and 2.8% system losses at peak output.
Has anyone done similar measurements?

No, that is not a system loss, it is a loss in the wiring between the inverter and the grid. I presume you measured the grid voltage, not the voltage at the inverter output which would mean nothing without knowing how the inverter performs unloaded. The inverter loss is the Pout/Pin and with modern designs can be as high as about 95%. That is not true with low voltage inverters, there most will be in the low 90's to high 80's.

7V X 17A is 119 watts dissipated in your wiring between the inverter and the grid!

[nerdralph]

No, that is not a system loss, it is a loss in the wiring between the inverter and the grid. I presume you measured the grid voltage, not the voltage at the inverter output which would mean nothing without knowing how the inverter performs unloaded. The inverter loss is the Pout/Pin and with modern designs can be as high as about 95%. That is not true with low voltage inverters, there most will be in the low 90's to high 80's.

7V X 17A is 119 watts dissipated in your wiring between the inverter and the grid!

Yes, as I said, the losses FROM my inverter TO the grid. To reduce these losses in other PV systems I build, I plan to increase the size of the wiring and/or breaker in the electrical panel.

[foggysail]

Yes, as I said, the losses FROM my inverter TO the grid. To reduce these losses in other PV systems I build, I plan to increase the size of the wiring and/or breaker in the electrical panel.

Yes, that is a wise move. You know the distance between your inverter and panel so you can use wire tables to calculate your expected loss before making a change.

[Ampster]

Yes, as I said, the losses FROM my inverter TO the grid. To reduce these losses in other PV systems I build, I plan to increase the size of the wiring and/or breaker in the electrical panel.

Just to clarify, if you upgrade the wire but have not increased the output rating of your inverter the breaker does not have to be upgraded. Also doing so may cause an inspector to miscalculate the backfeed rule for your bus bar loading. Traditionally the breaker is to protect the wiring so a smaller breaker on larger wire is acceptable.

[bcroe]

It is hard to use differential voltage measurements for these
calculations, unless they are made at the exact same instant.
Things are always shifting around. Part of a shift may be in
the PoCo wiring, which might be real loss, but that is their
loss, not yours.

I favor calculating the wire resistance from its length and gauge,
measure the current and use formulas like P = I x I x R

I have done such measurements, my dc losses over
hundreds of feet increase to a bit over 1% at full power, 400VDC.

My AC losses using previously in place 4 gauge wire, were in a
600 foot loop times .00025 ohm per foot = 0.15 ohm. 60A
gave 9V drop, 540W lost out of the inverter 15KW outpout. This
was excessive, and I eventually buried 4/0 aluminum wire,
to get AC wiring losses under 1%. Bruce Roe

[nerdralph]

It is hard to use differential voltage measurements for these
calculations, unless they are made at the exact same instant.
Things are always shifting around. Part of a shift may be in
the PoCo wiring, which might be real loss, but that is their
loss, not yours.

I favor calculating the wire resistance from its length and gauge,
measure the current and use formulas like P = I x I x R

I have done such measurements, my dc losses over
hundreds of feet increase to a bit over 1% at full power, 400VDC.

My AC losses using previously in place 4 gauge wire, were in a
600 foot loop times .00025 ohm per foot = 0.15 ohm. 60A
gave 9V drop, 540W lost out of the inverter 15KW outpout. This
was excessive, and I eventually buried 4/0 aluminum wire,
to get AC wiring losses under 1%. Bruce Roe

In my distribution zone, the variation from minute-to-minute is typically 0.1V - 0.2V. I keep a kill-a-watt plugged in, and throughout the day I rarely see a variation of more than 2V (121V-123V).

When I've calculated wire losses, I find it only accounts for about half of the resistance. The contacts and the breakers have resistance that has to be accounted for. You are correct that some of the losses I'm measuring are beyond the meter (between the meter and step-down transformer). I'd like to see the voltage being measured by my smart meter, but the utility only reports pseudo-realtime power consumption/production but not voltage. For a rainy day project I may setup logging on the voltage of the electrical panel bus. That would get me closer to measuring the losses up to the meter.

[nerdralph]

Just to clarify, if you upgrade the wire but have not increased the output rating of your inverter the breaker does not have to be upgraded. Also doing so may cause an inspector to miscalculate the backfeed rule for your bus bar loading. Traditionally the breaker is to protect the wiring so a smaller breaker on larger wire is acceptable.

Correct. I also find it's usually possible to upgrade the breaker without upgrading the wire. I suspect the NEC works similar to the CEC, which allows higher current on branch circuits based on the temperature rating of the terminal equipment. Electricians (and even inspectors) around here tend to go with the lowest 60C rating, even though breakers and the inverter connections tend to be rated for 90C. For 8/2 wiring, the default 60C rating limits the breaker to 40A, but with 90C you can use a 55A breaker (which will have lower resistance).

https://www.eaton.com/content/dam/ea...ap003001en.pdf

p.s. I tried uploading CEC Table 2, but I keep getting a "Error uploading image" message.

[bcroe]

p.s. I tried uploading CEC Table 2, but I keep getting a "Error uploading image" message.

I also tried to upload a couple times, pics I have used before, on this
Apple in use a long time, but got ERROR message. Bruce Roe