Single Phase Inverters on 208 3 Phase

[wwhitney]

Can someone show me a different situation in which L-L feeding a 3-phase wye transformer is as effective as effective as feeding phase to phase?

In the context of power from the grid defining all the voltage systems on the premises, a grid-tied inverter is equivalent to a negative load. In that you could add a single phase grid-tied inverter producing say 10 kW, connected L-L (or L-N), along with a single phase load consuming 10kW, connected L-L (or L-N), and the net result of adding both at once is that no other currents in the system will change. Equivalently, adding a 10kW output grid-tied inverter will have the same effect as deleting a 10kW load.

So would it be helpful to see examples in which a 3-phase transformer wye secondary can supply loads L-L as effectively as L-N?

Cheers, Wayne

 

[bellington]

In the context of power from the grid defining all the voltage systems on the premises, a grid-tied inverter is equivalent to a negative load. In that you could add a single phase grid-tied inverter producing say 10 kW, connected L-L (or L-N), along with a single phase load consuming 10kW, connected L-L (or L-N), and the net result of adding both at once is that no other currents in the system will change. Equivalently, adding a 10kW output grid-tied inverter will have the same effect as deleting a 10kW load.

Agreed that a grid tied inverter can supply the same power to a load that two legs of 208 can provide.

So would it be helpful to see examples in which a 3-phase transformer wye secondary can supply loads L-L as effectively as L-N?

No. I'm looking for any other situation that effectively provides power through a wye transformer connected leg to leg on the supply side.

Cheers, Wayne

 

[jaggedben]

Delta-wye transformers are used all the time. 480 delta to 208/120 is extremely common in commercial facilities, places like large hotels where there might be one for each floor. Not sure if that's exactly what you mean though.

 

[wwhitney]

No. I'm looking for any other situation that effectively provides power through a wye transformer connected leg to leg on the supply side.

OK, here's an example: take a 3P3W source and connect it to the wye side of a delta-wye transformer. The center point of the wye side will become a stable neutral point because the delta configuration ensures that the sum of the voltages to the center point of the wye will be 0. You could connect a L-N load to the wye side of the transformer and it would be efficiently powered by the delta source. Or you could power L-L loads from the delta side of the transformer.

Cheers, Wayne

 

[ggunn]

Can someone show me a different situation in which L-L feeding a 3-phase wye transformer is as effective as effective as feeding phase to phase?

I confess; I am lost here. Are you still trying to show that there is a major difference in efficiency between connecting three single phase inverters phase to phase (through a transformer or not) to a three phase service and connecting them phase to neutral? Once again, I can tell you from firsthand experience that you are mistaken as long as the voltages seen by the inverters are within their operating windows, though offhand I can't think of any inverters that can be connected either way on the same service. It makes no difference whether you understand it or not.

 

[bellington]

I discovered that I could change the winding ratio on the transformers in Falstad. This first circuit is a 208 wye connected L-L, with a 2.3-turn ratio transformer, to give 480/277 wye output across (3)100-Ω load resisters . It provides 4.4 kW per leg.

https://tinyurl.com/yvncfnt4

This one is a 240 delta, with a 2-turn ratio to give 480/277 wye output across (3) 100 Ω load resistors. It provides12.8 kW per leg.

https://tinyurl.com/ywtbhcdn

As two circuits that have no relationship to grid-tied inverters, what have I done wrong to get such a large difference in output across the same 100 Ω load resistors?

 

[bellington]

OK, here's an example: take a 3P3W source and connect it to the wye side of a delta-wye transformer. The center point of the wye side will become a stable neutral point because the delta configuration ensures that the sum of the voltages to the center point of the wye will be 0. You could connect a L-N load to the wye side of the transformer and it would be efficiently powered by the delta source. Or you could power L-L loads from the delta side of the transformer.

Cheers, Wayne

A delta transformer allows for current from from A into B and C, and continuing around the circle. A delta arrangement of inverters only allows current from A and through B, B through C, and C through A, and the reverse.

 

[wwhitney]

I discovered that I could change the winding ratio on the transformers in Falstad. This first circuit is a 208 wye connected L-L, with a 2.3-turn ratio transformer, to give 480/277 wye output across (3)100-Ω load resisters . It provides 4.4 kW per leg.

https://tinyurl.com/yvncfnt4

"Wye-connected L-L" makes no sense, you mean delta-connected L-L.

Your (3) 100Ω loads are delta connected on the 480Y/277V side, so they each dissipate V²/R = 480²/100 = 2.3 kW. That is the average power over a cycle; the peak instantaneous power, which you are noting by watching the watt meter, is twice that, or 4.6 kW. And your model isn't quite showing the full 4.6 kW, because you have those 250 mΩ resistors in series with your sources, so you are dropping some voltage there and not quite getting the full 480Y/277V on the left side. So you only see 4.4 kW peak, instead of 4.6 kW peak.

This one is a 240 delta, with a 2-turn ratio to give 480/277 wye output across (3) 100 Ω load resistors. It provides12.8 kW per leg.

https://tinyurl.com/ywtbhcdn

In this one you, you have 240V sources delta connected and going to the delta side of a transformer with a 2:1 turns ratio. But the other side of the transformer is wye-connected, so you have created a system with 480V L-N, i.e. 831Y/480V. With your (3) 100Ω loads still delta-connected, each one will dissipate 831²/100 = 6.9 kW average or 3 times the previous case.

what have I done wrong to get such a large difference in output across the same 100 Ω load resistors?

You switched from a wye-wye transformer in the first example to a delta-wye transformer in the second example. For the delta-wye transformer you'd need a turns ratio of 277/240; if you want to use a 2:1 turns ratio, you'd need a delta-delta transformer to make the left-hand voltage system match the first example. You could have noticed this problem by noticing that your load voltmeters in the second example are reaching over 1 kV peak.

Cheers, Wayne

 

[wwhitney]

A delta transformer allows for current from from A into B and C, and continuing around the circle. A delta arrangement of inverters only allows current from A and through B, B through C, and C through A, and the reverse.

Huh? I don't see any difference between your two descriptions, nor how it is relevant to the example I provided.

Cheers, Wayne

 

[bellington]

Sorry, I had incorrectly formed the primary 480/277 wye part of the circuits.

This should be the correct circuit of a 480/277 wye primary with (3) 30 Ω resistors connected N to L for 4.8 kW per leg. The (3) voltage sources are connected L-L (across 2 phases) on the 208/120 wye secondary. The transformer turns ratio is 2.3:

https://tinyurl.com/yv6k39jh

This should be the correct circuit of a 480/277 wye primary with (3) 30 Ω resistors connected N to L for 14.2 kW per leg. The (3) voltage sources are connected L-L (across 1 phase) on the 240 delta secondary. The transformer turns ratio is 2:

https://tinyurl.com/ywcmh63d

Viewing these circuits as totally separate from grid-tied inverters, are there still errors that would account for the huge difference in output?

 

[bellington]

Huh? I don't see any difference between your two descriptions, nor how it is relevant to the example I provided.

Cheers, Wayne

At a point in time in a circuit fed by a 3P3W circuit, there is current flowing into leg A and out of legs B and C. At that same point in time, an AB inverter can only flow into A and out of B, not out C.

 

[wwhitney]

This should be the correct circuit of a 480/277 wye primary with (3) 30 Ω resistors connected N to L for 14.2 kW per leg. The (3) voltage sources are connected L-L (across 1 phase) on the 240 delta secondary. The transformer turns ratio is 2:

https://tinyurl.com/ywcmh63d

I'm now calling the left-hand side of your examples the secondary, and the right hand side the primary, because your sources are on the right hand side. I forget if I've been consistent about this in my previous posts.

In this example you have the 240V sources connected L-N, and the transformer is wye-wye with a 2:1 turns ratio. So you are still making a 816Y/480V system on the secondary (left-hand) side.

To fix this, you could change the transformer turns ratio to 277/240 to go from your primary 416Y/240V system to a secondary 480Y/277V system. Or you could change your sources to be connected L-L, and preferably change your transformer to be delta-wye. If you change your sources to be L-L connected, and keep your transformer wye-wye, it will work as expected only if all the loads on both sides of the transformer are perfectly balanced, as your neutral point on both sides is floating.

Cheers, Wayne

 

[wwhitney]

At a point in time in a circuit fed by a 3P3W circuit, there is current flowing into leg A and out of legs B and C. At that same point in time, an AB inverter can only flow into A and out of B, not out C.

True for the AB inverter alone; one 2W single phase inverter obviously can't recreate 3P3W.

But if you have 3 delta-connected 2W single phase inverters connected AB, BC, and CA, then at the point in time in your 3P3W circuit when you expect "current flowing into leg A and out of legs B and C," the AB inverter can flow current into leg A from leg B, and the CA inverter can flow current into leg A from leg B. Which they will do if they are grid-following inverters synced to an existing 3 phase voltage system, just by following the existing voltages with their phase offsets.

Cheers, Wayne

 

[wwhitney]

So you are still making a 816Y/480V system on the secondary (left-hand) side.

As you adjust your examples, mentally draw a vertical line on your schematic going through your 3 transformers. If two examples have identical circuits on the left hand side of that line, but you find the left-hand side behavior doesn't match between the two examples, you can be 100% sure that your transformers are not making the same voltage system on the left-hand side. So if you were intending to make the same voltage system, you can be sure that you have a misconfiguration on the right hand side in one of the two examples.

Cheers, Wayne

 

[bellington]

I tried an AC source on both sides of a Y to Y 1:1 transformer with a 60 Ω resistor across each coil, just to get some current for comparison.

I often make careless mistakes, but this one should be 170-v peak L-N on the left (primary) to 294.2-v peak L-L on the right (secondary) with wattages rising to over 7kW:

https://tinyurl.com/ysjcyzvs

This one should be the exact same circuit with 170-v peak L-N on the left to 170-v peak L-N on the right with less than 500 W:

https://tinyurl.com/yw8zzcvz

What did I do wrong, or what am I missing, that makes it appear the L-N/L-L circuit is using almost 7kW, while the L-N/L-N circuits have minimum current?

 

[wwhitney]

I tried an AC source on both sides of a Y to Y 1:1 transformer with a 60 Ω resistor across each coil, just to get some current for comparison.

So that's potentially confusing, as you are putting two voltage sources in parallel. If you want to model that one of them is the grid, and one is a grid-following inverter, then to model that grid-following behavior you have to manually set the voltage and phase of one set of sources to match what they would see on their terminals from the other source. And then doing that doesn't really tell you anything, as the current required by the circuit will simply divide between the two voltages sources.

In other words, it's not really useful to have voltage sources on both sides of the transformer. Just keep your voltages sources on side of the transformer. You'd need falstad to implement current sources or voltage matching voltage sources to properly model grid-following inverters as sources.

I often make careless mistakes, but this one should be 170-v peak L-N on the left (primary) to 294.2-v peak L-L on the right (secondary) with wattages rising to over 7kW:

https://tinyurl.com/ysjcyzvs

Here you have a phase mismatch, the voltage sources on the right should be at the phases 330, 90, and 210 degrees from top to bottom to be in phase with the voltage sources on the left. If you change the phases to those values, the behavior matches the other example.

This one should be the exact same circuit with 170-v peak L-N on the left to 170-v peak L-N on the right with less than 500 W:

https://tinyurl.com/yw8zzcvz

Note that this circuit doesn't have any behavior that depends on the sources being 3 phase. Each pair of phase matched sources (or better yet, just delete one) can be set to an arbitrary phase, and it would still be the case that on each phase you will get 480W dissipated over your pair of 60 ohm resistors on each 120V source.

Cheers, Wayne

 

[bellington]

So the simulator is useless. If I use (3) single phase 240 volt inverters capable of 32 amps, or 7.6 kW, on 208, at 32 amps, I only get 6.6 kW. If I connect, the (3) inverters, capable of 7.6 kW, to 208, I have (3) coils with inverters only applying 120 volts to each of the (3) coils because each inverter is connected to (2) coils. Now the total kW per inverter becomes 120 X 32, 3.840 kW, or half of what the 7.6 kW. Seems like I need twice as many inverters on 208 than I would need on 240 to get the same capacity. Wouldn't I need 35 inverters for the 134 kW total capacity on 208, and only 18 on 240. Is that wrong?

 

[wwhitney]

If I use (3) single phase 240 volt inverters capable of 32 amps, or 7.6 kW, on 208, at 32 amps, I only get 6.6 kW.

Correct. And that is the only real downside to having a 208Y/120V system rather than a 240V delta system. Not enough of a downside to justify adding a transformer to create a 240V delta system. Adding 1-3 more inverters is better (cheaper/easier/more efficient) than adding a transformer, as I understand it (but I haven't actually done a cost comparison).

If I connect, the (3) inverters, capable of 7.6 kW, to 208, I have (3) coils with inverters only applying 120 volts to each of the (3) coils because each inverter is connected to (2) coils.

No, each 2-wire inverter is matching the 208V present across its terminals. There's no power connection from the inverter(s) to the neutral, so there's no sense in which the inverter is seeing or matching 120V. (*)

Now the total kW per inverter becomes 120 X 32, 3.840 kW, or half of what the 7.6 kW.

No, you've connected the inverter to 208V, not 120V, so you had it right when you came up with 6.6 kW.

Stop thinking about the transformer secondary configuration for a moment. If you have 3 wires with 208V between each pair of wires, and the phase shifts are 120 degrees apart, then you have a 3W Delta 208V (sub)system. If you attach grid-following inverters L-L to this system, each inverter will match the voltage it sees and put out 208V. It doesn't matter what the upstream transformer configuration is.

Cheers, Wayne

(*) Note that operationally, your single phase inverters may require a connection to the neutral, not because they are going to use it for power output, but just for instrumentation. I.e. they will check that you either have 240V L-L and 180 degree difference between L1-N and L2-N, or that you have 208V L-L and 120 degree difference between L1-N and L2-N, and refuse to put out power if they find something weird that doesn't match one of those two systems the inverter is programmed to interact with. But that is the only use for the neutral; for the purposes of power output and this discussion, we can ignore the neutral connection to the inverter.

The preceding paragraph is for typically available products. It's certainly theoretically possible to create an inverter that would connect to the neutral and use it for power output when the L-L voltage is 208V. In which case with a 32A current limit, it could put out a full 7.6 kW, by putting out 32A * 120V * 2, which would result in 32A on that neutral connection. But this is a hypothetical, I've not heard of any products that do that, although my knowledge of the field is primarily from monitoring this forum.

 

[ggunn]

So the simulator is useless. If I use (3) single phase 240 volt inverters capable of 32 amps, or 7.6 kW, on 208, at 32 amps only get 6.6 kW. If I connect, the (3) inverters, capable of 7.6 kW, to 208, I have (3) coils with inverters only applying 120 volts to each of the (3) coils because each inverter is connected to (2) coils. Now the total kW per inverter becomes 120 X 32, 3.840 kW, or half of what the 7.6 kW. Seems like I need twice as many inverters on 208 than I would need on 240 to get the same capacity. Wouldn't I need 35 inverters for the 134 kW total capacity on 208, and only 18 on 240. Is that wrong?

It's going to be hard if not impossible to get an apples to apples comparison between three inverters connected phase to phase and the same inverters connected phase to neutral because there are no single phase inverters that I know of that can connect to either 120V or 208 or to either 480V or 277V.

All I can say is that if you have three inverters that produce 10kVA each at 120V and three other inverters that produce 10kVA each at 208V, and you connect the 120V inverters phase to neutral on a 208/120V service and the 208V inverters phase to phase, both scenarios will deliver a total of 30kVA.

That said, I have not seen single phase inverters connected phase to neutral to a three phase service in many years, not since the old transformer coupled SMA inverters that could connect at either 240V or 277V.

 

[ruxton.stanislaw]

That said, I have not seen single phase inverters connected phase to neutral to a three phase service in many years, not since the old transformer coupled SMA inverters that could connect at either 240V or 277V.

I might be a bit out of my depths here, but the Growatt WIT 100K-HU-US (three phase inverter, configured as either 3P3W+PE or 3P4W+PE) can be connected with 100% imbalance (i.e. one live conductor and neutral). The default voltage is 277 V, but the EPS voltage can be reconfigured to 127 V with the default firmware. I am assuming it has reduced power output at the lower voltage.