Single Phase Inverters on 208 3 Phase

[bellington]

I agree.

I also agree.

This is where I don't agree, at least for one wave passing through two phase angle legs of a transformer's secondary coils. If we are using our resultant wave and sending it back through A-B, then we are pushing electrons backwards through B-N most of the time.

Sorry, it would be more like both legs would have electrons pushed the wrong way during 60 degrees for each one.

 

[wwhitney]

The exception to this statement is that your purple resultant waveform, even reversing the polarity of measurement of B-N to N-B on your graph, the resultant wave still has 60 degrees of time that the resultant wave being fed back into the secondary is pushing electrons in the opposite direction of A-N and another 60 degrees of time the resultant wave is pushing electrons in the opposite direction of N-B.

This is more or less correct. If we think of the inverter connected A-B as pushing current back through A-N and N-B, then yes, for 1/6 of each cycle, the current it pushes though A-N will not be carrying energy back upstream but the energy flow will instead be like a load. Same for N-B, although a different 1/6 of the cycle.

But all that means is this setup uses a bit more current to transfer power than a setup with individual A-N and B-N inverters. The net power transfer through the A-N coil will be 86.6% as much as it would be if the current through the A-N coil were in phase with the voltage (with a negative sign for power backfeed).

This 100% vs 86.6% does not represent a loss of power generated.

The primary side of the transformer doesn't replicate this battle, only the results.

The primary side coils will always see current and voltage in exactly the same phase relationship to each other as in the secondary coils. That's how transformers work.

What you see outside the transformer depends on the transformer configuration. If it is wye-wye, then you'll see a current between primary lines A' and B' exactly comparable to the current the inverter is pushing out between B and A. The "battle" will be fully replicated. (Not a battle, just a power factor less than 1).

But if the transformer is delta-wye, the story is more complicated, and I think (not 100% sure) there would be current on A', B' and C'.

Those results feeding back through the primary will be less than what we put into the secondary.

Some quantities will be less or will be more, but power transferred will still be the same.

Cheers, Wayne

 

[jaggedben]

The exception to this statement is that your purple resultant waveform, even reversing the polarity of measurement of B-N to N-B on your graph, the resultant wave still has 60 degrees of time that the resultant wave being fed back into the secondary is pushing electrons in the opposite direction of A-N and another 60 degrees of time the resultant wave is pushing electrons in the opposite direction of N-B. The primary side of the transformer doesn't replicate this battle, only the results. Those results feeding back through the primary will be less than what we put into the secondary.

In your bolded statement you are expecting the transformer to do something which you've been repeatedly been told is not done by the transformer. It's done by the 'grid', i.e. the BESS. The BESS is what 'replicates the battle'. It does not matter if there is a transformer in between the source of the single wave and the BESS, or not.

Say your BESS/grid was simply 208/120 wye. No transformers. Your question all along still stands: how does a multi-phase BESS charge from a solar inverter that is outputting a single phase? It does it by charging with two inverter/chargers, in two waves that are phase shifted from each other. The trigonometric addition/subtraction of current is the same whether or not the two waves are the source and the single wave is the load, or whether the single wave is the source and the two waves are the load. As Wayne tried to walk you through, the current adds up in either direction; there is no loss in one direction that doesn't occur in the other; if 1+1=2 then 2=1+1; there is no loss because I reversed the direction I typed that. Go back to the graph again if you need the visual to confirm this.

Got that?

Okay then, the transformer doesn't fundamentally change this. (Especially if you have a wye-wye, as you seemed to imply above.) It passes through all the waveforms that come to it. If the BESS can 'break apart' a single waveform on the secondary side of the transformer, then a BESS can do it on the primary side, too. So no, the purple line is not an exception to the statement I made about before and after the transformer. The BESS will still be able to find the single waveform and break it apart on the other side of the transformer.

 

[Carultch]

Sorry, it would be more like both legs would have electrons pushed the wrong way during 60 degrees for each one.

Part of the cycle they are pushed the wrong way, but most of the cycle, the electrons are moving the correct way. Power factor is a measure of how effective the two waveforms interact with each other to transfer power.

The whole idea of using angles and vector geometry to keep track of AC waveforms, is really a mathematical abstraction that makes it easier to see visually how the numbers relate to each other. There are no real world distances to measure with a ruler or angles to measure with a protractor, on an actual circuit, that have anything to do with this vector math. It is a scale model of amplitudes of waveforms, where cosine amplitude is plotted on one axis, and sine amplitude is plotted on the other axis.

What is really happening, is a product of out-of-phase voltage and current waveforms to calculate power, and an integral that calculates the average power. The apparent power (units of VA) is the product of RMS amplitudes of voltage and current waveforms, ignoring the fact that they are out of phase. The real power is the power that accounts for instantaneous power at every point along the way, and averages that instantaneous power. This (real power calculation) is what is known as an inner product of two functions, which has properties in common with a dot product of vectors.

Side note: "RMS amplitudes" are a topic for another day, but the short answer is that it is a special kind of time averaging, that for an ideal waveform, is actual amplitude divided by sqrt(2). The nominal voltages and currents you see all the time, are really RMS amplitudes rather than actual amplitudes.

 

[bellington]

This is more or less correct. If we think of the inverter connected A-B as pushing current back through A-N and N-B, then yes, for 1/6 of each cycle, the current it pushes though A-N will not be carrying energy back upstream but the energy flow will instead be like a load. Same for N-B, although a different 1/6 of the cycle.

But all that means is this setup uses a bit more current to transfer power than a setup with individual A-N and B-N inverters. The net power transfer through the A-N coil will be 86.6% as much as it would be if the current through the A-N coil were in phase with the voltage (with a negative sign for power backfeed).

This 100% vs 86.6% does not represent a loss of power generated.


The primary side coils will always see current and voltage in exactly the same phase relationship to each other as in the secondary. That's how transformers work.

What you see outside the transformer depends on the transformer configuration. If it is wye-wye, then you'll see a current between primary lines A' and B' exactly comparable to the current the inverter is pushing out between B and A. The "battle" will be fully replicated. (Not a battle, just a power factor less than 1).

But if the transformer is delta-wye, the story is more complicated, and I think (not 100% sure) there would be current on A', B' and C'.

Some quantities will be less or will be more, but power transferred will still be the same.

Cheers, Wayne

This is more or less correct. If we think of the inverter connected A-B as pushing current back through A-N and N-B, then yes, for 1/6 of each cycle, the current it pushes though A-N will not be carrying energy back upstream but the energy flow will instead be like a load. Same for N-B, although a different 1/6 of the cycle.

But all that means is this setup uses a bit more current to transfer power than a setup with individual A-N and B-N inverters. The net power transfer through the A-N coil will be 86.6% as much as it would be if the current through the A-N coil were in phase with the voltage (with a negative sign for power backfeed).

This 100% vs 86.6% does not represent a loss of power generated.

Does it not represent 5.7 kW on the primary side for the 6.6 kW I put in on the secondary?

The primary side coils will always see current and voltage in exactly the same phase relationship to each other as in the secondary coils. That's how transformers work.

What you see outside the transformer depends on the transformer configuration. If it is wye-wye, then you'll see a current between primary lines A' and B' exactly comparable to the current the inverter is pushing out between B and A. The "battle" will be fully replicated. (Not a battle, just a power factor less than 1).

But if the transformer is delta-wye, the story is more complicated, and I think (not 100% sure) there would be current on A', B' and C'.

Some quantities will be less or will be more, but power transferred will still be the same.

Cheers, Wayne

 

[ggunn]

That has been my concern. My head has no problem using a single phase inverter connected to a single phase leg of a transformer secondary, be it 240 delta, or 416/240Y. My head does have issues with a single phase inverter attempting to back feed two phases.

Before smaller three phase inverters were widely available (by cracky!) I designed many systems that connected single phase inverters to three phase services. Some connected phase to phase and some connected phase to neutral. The phase angle is only relevant if the neutral is part of a circuit with two of the phases or if all three phases are involved.

 

[bellington]

Before smaller three phase inverters were widely available (by cracky!) I designed many systems that connected single phase inverters to three phase services. Some connected phase to phase and some connected phase to neutral. The phase angle is only relevant if the neutral is part of a circuit with two of the phases or if all three phases are involved.

I'm fine with single phase inverters connected to a single phase, just not connected to two legs.

 

[ggunn]

Does it not represent 5.7 kW on the primary side for the 6.6 kW I put in on the secondary?

No. For a transformer, ignoring losses which are typically relatively small, by the Law of Conservation of Energy VA in = VA out. Start from there.

 

[ggunn]

I'm fine with single phase inverters connected to a single phase, just not connected to two legs.

It doesn't matter if you are fine with it or not. It works. I have connected inverters this way many times.

 

[bellington]

It doesn't matter if you are fine with it or not. It works. I have connected inverters this way many times.

Do you currently have a system running that is connected 3 single phase inverters to each pair of legs on a 208 Y system on which you can show me the efficiency?

 

[ggunn]

Do you currently have a system running that is connected 3 single phase inverters to each pair of legs on a 208 Y system on which you can show me the efficiency?

Not at my fingertips, but as I said VA in must (virtually) equal VA out for a transformer no matter how it is configured. What we are saying is readily verifiable once you wrap your head around how three phase power works. One thing that at first seems counterintuitive is that if you measure V to N for each of the phases, the waveforms are 120 degrees apart, but if you measure Va to Vb you see a single waveform that looks like any other single phase with 180 degrees between phases.

 

[wwhitney]

Does it not represent 5.7 kW on the primary side for the 6.6 kW I put in on the secondary?

No. Let's try this:

We've established that the A-B voltage is 208V, and it is 30 degrees out of phase with the N-B voltage of 120V as well as the A-N voltage of 120V, yes?

So we have a 208V single phase inverter installed A-B, and we have a simple circuit of inverter, N-B coil, A-N coil. Our inverter puts out 10A in phase with A-B. 10A * 208V = 2080W of power pushed out.

In our simple circuit, with 10A through the inverter, we must also have 10A through N-B and 10A through A-N. [We can't gain or lose current in this circuit, it's just a loop and the electrons aren't piling up at any place on the loop.] And 10A * 120V = 1200W through N-B, likewise 1200W through A-N, for a total of 2400W. Where did we magically get 320W from?

The answer is we didn't. In AC, Watts = Volts * Amps only when the voltage and the current are in phase. Which is true for the inverter A-B, but not true for the coils N-B and A-N. In those coils, the voltage and the current are 30 degrees out of phase (since the current is in phase with the voltage A-B). The AC formula for this case is Watts = Volts * Amp * cosine(phase difference). Cosine(30 degrees) = 0.866. . . So for coil N-B, we have Watts = 120 * 10 * 0.866 = 1040W. Likewise for coil A-N, 1040W. And as expected, the power through both coils matches the power put out by the inverter, 1040W + 1040W = 2080W.

Compare this to two 120V inverters, one connected A-N and one connected B-N. For them to put out 2080W total, they'd have to each put out 1040W. At 120V, with current in phase with voltage, that would require 1040/120 = 8.66A. Now we have two separate loops, the first the inverter A-N and the coil N-A; the second the inverter B-N and the coil N-B. Here the current in each loop is 8.66A, and each voltage in each loop is in phase with the current in that loop. But unlike the first case, the currents in the two coils are not in phase.

So the difference between the two cases is that the 208V inverter needs to use 10A through both coils to deliver 2080W, while the two 120V inverters can use just 8.66A through each coil to deliver 2080W. But in both cases all the power put out by the inverter makes it to the transformer coils, without any losses.

Which doesn't actually answer the quoted question above, but I hope speaks to what you are having trouble seeing. The transformer similarly transfers all the power from the primary side to the secondary side (or vice versa) without any power losses due to power factor issues like this. [That factor, cosine(30 degrees) = 0.866. . . , is called the power factor.]

Cheers, Wayne

 

[Carultch]

Do you currently have a system running that is connected 3 single phase inverters to each pair of legs on a 208 Y system on which you can show me the efficiency?

One thing I think might be confusing you, is the difference between efficiency and power factor. They are both measured as a percent, and in an ideal case, they both equal 100%. But they are not the same thing.

Efficiency refers the ratio of output energy, to input energy. The implication of this term, is that any difference is lost energy, due to heat generation. The energy still exists, but it leaves the electrical domain and enters the thermal domain.

Power factor refers to the ratio of the more accurate "product" of volts and amps that matters (i.e. real power), to the apparent product of volts and amps (i.e. apparent power). The reason it isn't always 100%, is that the waveforms aren't synchronized. The energy still remains in the electrical domain, it is just time-delayed.

The power characteristics of an interphase 208V inverter are a case where they produce a different power factor when comparing to the current on an individual phase conductor, to the power factor the inverter actually produces. In a balanced group, the phase shifts all add up to zero, and three interphase inverters produce no different of a result, than had you installed three 120V inverters of the same power connected across individual phases to neutral. An individual interphase inverter may have a different product of phase-to-neutral voltage and current that adds up to more than its power rating, but if you account for phase shifts between them, you'll see that no actual power is lost. Apparent power may appear to be lost, but apparent power isn't power. Apparent power is a concept for infrastructure sizing, to account for the ineffectiveness of phase shifts that make more current than would be needed.

 

[ggunn]

One thing I think might be confusing you, is the difference between efficiency and power factor.

From what I am reading, I don't think this is the case. I think that the root of his difficulty is a lack of understanding of three phase systems.

 

[jaggedben]

Do you currently have a system running that is connected 3 single phase inverters to each pair of legs on a 208 Y system on which you can show me the efficiency?

Not at my fingertips...

@ggunn Haven't you in the past done systems with single-phase inverters connected L-L to 208/120 services? On any of those systems did the customer complain that energy production or net-metering savings were 13% less than expected?

For that matter, I believe any of the systems where you connected three-phase wye inverters through a delta-wye step-up also count to this question.

 

[ggunn]

@ggunn Haven't you in the past done systems with single-phase inverters connected L-L to 208/120 services? On any of those systems did the customer complain that energy production or net-metering savings were 13% less than expected?

Yes, many times to the first; no to the second.

For that matter, I believe any of the systems where you connected three-phase wye inverters through a delta-wye step-up also count to this question.

OK, but the answer is still no. In either case VA in = VA out.

 

[bellington]

No. Let's try this:

We've established that the A-B voltage is 208V, and it is 30 degrees out of phase with the N-B voltage of 120V as well as the A-N voltage of 120V, yes?

So we have a 208V single phase inverter installed A-B, and we have a simple circuit of inverter, N-B coil, A-N coil. Our inverter puts out 10A in phase with A-B. 10A * 208V = 2080W of power pushed out.

In our simple circuit, with 10A through the inverter, we must also have 10A through N-B and 10A through A-N. [We can't gain or lose current in this circuit, it's just a loop and the electrons aren't piling up at any place on the loop.] And 10A * 120V = 1200W through N-B, likewise 1200W through A-N, for a total of 2400W. Where did we magically get 320W from?

The answer is we didn't. In AC, Watts = Volts * Amps only when the voltage and the current are in phase. Which is true for the inverter A-B, but not true for the coils N-B and A-N. In those coils, the voltage and the current are 30 degrees out of phase (since the current is in phase with the voltage A-B). The AC formula for this case is Watts = Volts * Amp * cosine(phase difference). Cosine(30 degrees) = 0.866. . . So for coil N-B, we have Watts = 120 * 10 * 0.866 = 1040W. Likewise for coil A-N, 1040W. And as expected, the power through both coils matches the power put out by the inverter, 1040W + 1040W = 2080W.

Compare this to two 120V inverters, one connected A-N and one connected B-N. For them to put out 2080W total, they'd have to each put out 1040W. At 120V, with current in phase with voltage, that would require 1040/120 = 8.66A. Now we have two separate loops, the first the inverter A-N and the coil N-A; the second the inverter B-N and the coil N-B. Here the current in each loop is 8.66A, and each voltage in each loop is in phase with the current in that loop. But unlike the first case, the currents in the two coils are not in phase.

So the difference between the two cases is that the 208V inverter needs to use 10A through both coils to deliver 2080W, while the two 120V inverters can use just 8.66A through each coil to deliver 2080W. But in both cases all the power put out by the inverter makes it to the transformer coils, without any losses.

Which doesn't actually answer the quoted question above, but I hope speaks to what you are having trouble seeing. The transformer similarly transfers all the power from the primary side to the secondary side (or vice versa) without any power losses due to power factor issues like this. [That factor, cosine(30 degrees) = 0.866. . . , is called the power factor.]

Cheers, Wayne

Thank you. Maybe that helped a little. If I had an inverter firing at 180 degrees out of phase with the resultant wave that a 208 load would see, I get no current, but also no power lost. With emphasis from ggunn of VA in = VA out, even though the 208 resultant wave has those period of time where they cancel each, the electrons aren't really in a battle, they just don't move.

 

[jaggedben]

Yes, many times to the first; no to the second.

OK, but the answer is still no. In either case VA in = VA out.

@bellington take note. People on this forum have done it in the real world with no energy lost.

 

[wwhitney]

If I had an inverter firing at 180 degrees out of phase with the resultant wave that a 208 load would see, I get no current, but also no power lost.

That's not correct. Your 208V inverter does push out current 180 degrees out of phase with what a 208 load would draw. So that if you connect a 208V resistive load in parallel with your inverter, and the load wants to draw the same current that your inverter is pushing out, those two currents add up to zero from the point of view of the main voltage source.

Cheers, Wayne

 

[jaggedben]

Does it not represent 5.7 kW on the primary side for the 6.6 kW I put in on the secondary?

To boil down Wayne's longer explanation...

No because 6.6kW at 208V at 100% power factor is the same amps as two times 3.3kW at 120V at 86.6% power factor.

6600/208*1=31.73A This is both your inverter circuit amp measurement and your amp measurements on A-N and B-N on the feeder if all the power is backfeeding to the grid.

31.73A*120V*.866= 3297 That's your watts per L-N phase.
times two L-N phases = 6594 watts The difference is just due to rounding off (mostly that 120V times the square root of 3 is actually 207.846..V)