Single Phase Inverters on 208 3 Phase

[jaggedben]

Just to be concrete here, if the inverter's maximum continuous output current is 32A, then putting (3) of these in a delta configuration directly on a single 3-pole 70A breaker (70A = 32A * 125% * sqrt(3)) is not an option? You'd need to use (3) double pole 40A breakers?

Conceptually, this can be done with some micro-inverter systems at lower amperage.

I guess even if it isn't required by the NEC or the manufacturer's instructions, using (3) double pole 40A breakers would be more convenient for maintenance and troubleshooting, so you could turn off just one inverter.

Each inverter requires its own isolating device per code, so with non-micro inverters that generally leads to each having its own 2-pole breaker.

 

[bellington]

Thank you. Yes, understood the generator concept requires an existing voltage to match. And even though 3 and 4 breakers are still connected to the same B leg inside the box, using another double breaker provides adequate isolation for each inverter. Conceptually, two wires on each breaker gets my head right for the delta configuration. Practicality for service, each inverter needs to be on its own double breaker. I'm wasting breaker space, but providing clarity for service.

Thanks

 

[synchro]

...
Second, on this question of delta connection on a wye secondary, there is nothing special about inverters, it is the same with loads. We use delta connected loads on a voltage system supplied with a wye transformer secondary all the time. Both 3 phase motors and groups of three 2-wire loads in a delta arrangement. It works efficiently.

Again, thanks for the patience and time spent. If I am any closer to grasping the single phase inverters becoming a 3-phase generator, the delta configuration for the inverters connected to the wye secondary was the key.

The phrase "3-phase generator" is a bit of a misnomer here, as it suggests that (3) of these inverters could be used to create a stand-alone 3-phase grid. Which they can't as grid-tied inverters. They rely on the existing 3-phase grid for all the of "3-phaseness" of what they are doing. They just play follow the leader--each one just sees a single compliant voltage waveform and then pushes out some current in phase with that waveform (for the default case where they are configured to have power factor 1).

I concur with the points that Wayne has made here. The voltages and the phase shifts between them are established by the grid when using grid-tied inverters. The inverters supply current to the grid at whatever voltage that might be (as long as it's within an accepted range).

The current drawn by a load resistance R is of course I = V / R via Ohm's Law. A grid-tied inverter equivalently presents a negative resistance such that the polarity of the current is reversed from the direction with a load, but otherwise all of the vector calculations for L-L loads on a 3-phase wye secondary will apply. And so the currents from the inverters can be treated just like load currents are handled.

 

[PWDickerson]

Conceptually I think you are on the right track, but you would not connect two inverters to the same breaker, and your numbering is off for a conventional 3 phase panel.

Here's a breaker space diagram color (found with google) coded by leg for a typical panel:

https://www.pinterest.jp/pin/7459155627111786/

So for your first group of three inverters, the first would be in spaces 1&3, the second 4&6, and the third 3&7. The next group would start with 8&10, and so on.

An engineer could calculate the line currents for you. Your BESS manufacturer ought to have support engineers who can tell you how important it is to have multiples of 3.

Should be 1&3, 4&6, and 5&7

 

[bellington]

Sorry, all was good until I started thinking. Here's where my brain has gone. Help me see the error. When the inverter for phase A shoots one complete cycle of 208 volts through legs A and B of the Y secondary, if it is timed to phase A, the current going through phase B 120 volt leg is 120 degrees ahead phase B. The only current correctly timed and applied is to the 120 volt coil of the A leg of the Y secondary. Only that A coil sends the correct energy to the 277 volt A coil in the Y primary of the transformer. I have effectively LOST most of the current passing through the B leg.

If that's wrong, please help me get past it.

 

[wwhitney]

Sorry, all was good until I started thinking. Here's where my brain has gone. Help me see the error. When the inverter for phase A shoots one complete cycle of 208 volts through legs A and B of the Y secondary, if it is timed to phase A,

It is not timed to phase A, if by saying phase A you mean the voltage waveform between A and N. It is timed to (in phase with) the voltage waveform between A and B.

the current going through phase B 120 volt leg is 120 degrees ahead phase B. The only current correctly timed and applied is to the 120 volt coil of the A leg of the Y secondary.

I'm not quite sure what you mean here, but current does not have to be in phase with voltage for energy transfer to occur. If the current is a full 90 degrees out of phase with the voltage, then there will be no energy transfer. If you compare the energy transfer for a fixed current as the phase angle changes from 0 degrees to 90 degrees, then the energy transfer will decrease from its maximum at 0 degrees to 0 energy transfer at 90 degres.

Only that A coil sends the correct energy to the 277 volt A coil in the Y primary of the transformer.

It is unlikely that your 480V : 208Y/120V transformer has a wye primary. Most likely it has a delta primary.

With the corrections out of the way, I'm pretty sure the following is correct, and it may be the picture you want:

The current the inverter puts out in phase with the voltage A-B will be 30 degrees out of phase with the voltage A-N and 30 degrees out of phase with the voltage N-B. So that conveys energy back to the transformer secondary using more current for a given rate of power transfer (or less power for a given current) than you'd have with 2 separate 120V inverters installed A-N and B-N, where the current-voltage phase angle would always be 0. But the energy still all gets transfer, there's no first order loss from using a bit more current, slightly out of phase with the L-N voltages.

This comparison is entirely analogous to your comparison of (2) 10A 120V hair dryers vs (1) 10A 208V hair dryers. They both "use" 10A on A-N and B-N, but the first case transfers 2400W, while the second transfers 2080W. But there's no sense of "loss" of 320W; in the second case the upstream sources never delivered the 320W in the first place.

Cheers, Wayne

 

[jaggedben]

Sorry, all was good until I started thinking. Here's where my brain has gone. Help me see the error. When the inverter for phase A shoots one complete cycle of 208 volts through legs A and B of the Y secondary, if it is timed to phase A, the current going through phase B 120 volt leg is 120 degrees ahead phase B. The only current correctly timed and applied is to the 120 volt coil of the A leg of the Y secondary. Only that A coil sends the correct energy to the 277 volt A coil in the Y primary of the transformer. I have effectively LOST most of the current passing through the B leg.

If that's wrong, please help me get past it.

Yes it's wrong, the 208V A-B sine wave is not in phase with either A-N or B-N. However if you had another inverter in A-C then the vector sum of their injection of current at terminal A *would* be in phase with the A-N waveform. And if you didn't have that second inverter, then the rest of the current for the A-N loads would come from the grid.

(I'm oversimplifying by assuming unity power factor everywhere and not considering the scenario where the load is less than the output of the inverters.)

 

[bellington]

It is not timed to phase A, if by saying phase A you mean the voltage waveform between A and N. It is timed to (in phase with) the voltage waveform between A and B.

Cheers, Wayne

It seems to me the resultant waveform of A-B as seen by a water heater would be a waveform of 208 volts, seen by the load as 60 degrees after phase A and 60 degrees before phase B. If I inject that same resultant wave, 208 volts at 60 degrees after A and 60 degrees before B, into the two A-B coils of the Y secondary, they can't subtract the individual desired waves from the one.

 

[wwhitney]

It seems to me the resultant waveform of A-B as seen by a water heater would be a waveform of 208 volts, seen by the load as 60 degrees after phase A and 60 degrees before phase B.

If we say the A-N voltage waveform peaks (maximum) at 0 degrees, and that the B-N voltage waveform peaks at 120 degrees, then the A-B voltage waveform peaks at -30 degrees. (Here I am being careful about direction, and "A-N voltage" means the voltage from N to A.)

If I inject that same resultant wave, 208 volts at 60 degrees after A and 60 degrees before B, into the two A-B coils of the Y secondary, they can't subtract the individual desired waves from the one.

I don't understand what that means. Who is "they"?

Say we restrict out consideration to a portion of the transformer secondary A-N-B and just one inverter or load connected A-B. There is only one current waveform in the loop A-N coil to B-N coil to A-B load/inverter back to A-N coil (here I'm not being careful about direction, e.g. maybe I should say A-N coil to N-B coil, depending on my direction conventions). That current waveform will be in phase with the A-B voltage (as a positive number for a load, and a negative number for an inverter) but 30 degrees out of phase (again up to sign since I wasn't careful with directions) with the A-N voltage waveform and the B-N voltage waveform.

When the current I through a coil is 30 degrees out of phase with the voltage V across that coil, the power transfer is |I| * |V| * cos(30 degrees). You would say the power factor is cos(30 degrees) = sqrt(3)/2 = 0.866 . . .

Cheers, Wayne

 

[jim dungar]

It seems to me the resultant waveform of A-B as seen by a water heater would be a waveform of 208 volts, seen by the load as 60 degrees after phase A and 60 degrees before phase B. If I inject that same resultant wave, 208 volts at 60 degrees after A and 60 degrees before B, into the two A-B coils of the Y secondary, they can't subtract the individual desired waves from the one.

You may need to read up on vector analysis. Way back in my early engineering classes we spent 34 weeks studying 3-phase circuits and only 2 weeks on 1-phase. When performing circuit analysis, 1-phase systems are basically nothing more than 3-phase with one vector set to a length of zero.

 

[bellington]

If we say the A-N voltage waveform peaks (maximum) at 0 degrees, and that the B-N voltage waveform peaks at 120 degrees, then the A-B voltage waveform peaks at -30 degrees. (Here I am being careful about direction, and "A-N voltage" means the voltage from N to A.)


I don't understand what that means. Who is "they"?


Cheers, Wayne

If A-N voltage peaks at 0 degrees and 120 degrees later B-N peaks, then the resultant waveform seems logically to be halfway between the two at 60 degrees. How do you get -30 for the resultant peak voltage for A-B?

"They' are coils A-N and B-N.

 

[bellington]

You may need to read up on vector analysis. Way back in my early engineering classes we spent 34 weeks studying 3-phase circuits and only 2 weeks on 1-phase. When performing circuit analysis, 1-phase systems are basically nothing more than 3-phase with one vector set to a length of zero.

So what is the resultant wave?

 

[wwhitney]

If A-N voltage peaks at 0 degrees and 120 degrees later B-N peaks, then the resultant waveform seems logically to be halfway between the two at 60 degrees. How do you get -30 for the resultant peak voltage for A-B?

If we write Vxy for the voltage from X to Y (the lower case letters should be read as subscripts), then we have:
Vab = Vnb + Van = Van - Vbn
since reversing the order introduces a minus sign.

Your statement is correct that the simple sum of waveforms will peak at the average of the two phase angle peaks of the two waveforms. But that means you've calculated the peak of Van + Vbn, while we need the peak of Van + (-Vbn).

If Vbn peaks at 120 degrees, then the peak of (-Vbn) is 180 degrees later, or at (120 + 180) = 300 = -60 degrees. And now when we sum (-Vbn) and Van, we can average the peak angles of 0 degrees and -60 degrees to get -30 degrees as the peak of Vab = Van - Vbn.

An important point of view I've omitted so far, because it is easier to type than to draw pictures, is that all this math can be represented by plane geometry. Call the origin N. Then if |Van| = 120V, A is represented by the point (120,0), and Van is an arrow from N to A, i.e. an arrow pointing to the right of length 120. Similarly Vbn is just that same arrow rotated 120 degrees around the origin N, which determines the point B. And Vab is just the arrow from B to A. That arrow has an angle of 30 degrees below the x-axis (and 30 degrees below Van).

Cheers, Wayne

 

[jaggedben]

If A-N voltage peaks at 0 degrees and 120 degrees later B-N peaks, then the resultant waveform seems logically to be halfway between the two at 60 degrees. How do you get -30 for the resultant peak voltage for A-B?

"They' are coils A-N and B-N.

When you measure the voltage A-B you are measuring the series voltage A-N-B. Thus you need to compare A-B to A-N and N-B (not B-N!) to see the phase relationship on a graph. N-B being the negative (inverse) of B-N is like flipping B-N upside down on the graph. When you do this you will find that the positive peaks of A-N and N-B are 60 degs apart and the peak of (A-N)+(N-B) is right in the middle 30 degs apart from each.

There is an opposite procedure in which you'd take (for example) A-B and C-A, flip C-A upside down to A-C, and find the peak of A-N 30 degs between the peak of the others.

Perhaps I can post a visual later of wxactly what I mean, but in the meantime try looking at the graph in this post from Carultch that I found with google.

208V 2 Pole Load

A load of 10kw on A and B phases has 10kw load on each phase not 5 kw on each. That is why we get 48 amps on A and B phases. Now we add a load to B and C phase and we still have 48 on Phase A but how do we only have 83 amps on B phase. Why not 96. How do we get 83 amps. Say you have a...

forums.mikeholt.com

 

[winnie]

If A-N voltage peaks at 0 degrees and 120 degrees later B-N peaks, then the resultant waveform seems logically to be halfway between the two at 60 degrees. How do you get -30 for the resultant peak voltage for A-B?

The above analysis suggests that you are still thinking of the voltage _at_ A and _at_ B.

Voltage cannot be defined _at_ a single point. It is always measured between two points. This is why you have to explicitly say the voltage from A to N or the voltage from B to N. You need to provide both points in the measurement.

The voltage seen by the inverter is the voltage from A to B. Not the average of A and B.

For sine wave AC voltage, a voltage may be represented as a two dimensional vector, where the magnitude of the vector represents the amplitude (voltage) of the sine wave, and the direction of the vector represents the phase.

If the A to N voltage is called Avec, and the B to N voltage is called Bvec, then the A to B voltage is Avec - Bvec using vector math. The 30 degree phase angles described by others come from doing this calculation.

-Jon

 

[jim dungar]

T

So what is the resultant wave?

The resultant wave of what?
Vector analysis does not use waves.

 

[wwhitney]

Vector analysis does not use waves.

To be fair, the individual vectors may be considered sine waves of the form f(t) = A cos(w(t-phi)), where w is a fixed cyclic frequency (w = 2*pi*60 for 60 Hz), and A (amplitude) and phi (phase shift) are free to vary. The advantage of vector analysis is that adding and scaling such functions is equivalent to just dealing with two real numbers (or one complex number), by associating to f the pair (A cos(phi), A sin(phi)) (or the complex number A * exp(i * phi)).

Cheers, Wayne

 

[bellington]

When you measure the voltage A-B you are measuring the series voltage A-N-B. Thus you need to compare A-B to A-N and N-B (not B-N!) to see the phase relationship on a graph. N-B being the negative (inverse) of B-N is like flipping B-N upside down on the graph. When you do this you will find that the positive peaks of A-N and N-B are 60 degs apart and the peak of (A-N)+(N-B) is right in the middle 30 degs apart from each.

There is an opposite procedure in which you'd take (for example) A-B and C-A, flip C-A upside down to A-C, and find the peak of A-N 30 degs between the peak of the others.

Perhaps I can post a visual later of wxactly what I mean, but in the meantime try looking at the graph in this post from Carultch that I found with google.

208V 2 Pole Load

A load of 10kw on A and B phases has 10kw load on each phase not 5 kw on each. That is why we get 48 amps on A and B phases. Now we add a load to B and C phase and we still have 48 on Phase A but how do we only have 83 amps on B phase. Why not 96. How do we get 83 amps. Say you have a...

forums.mikeholt.com

The graph did not help. My brain says if we start at 0 volts A-N,
at 0.0041667 of a second A-N reaches + peak,
at 0.0055556 of a second B-N reaches 0,
at 0.0083333 of a second A-N reaches 0,
at 0.0097222 of a second, B-N reaches + peak.

How can the results of two wave peaks that happened between time 0 and .0097222 of a second create a resulting wave + peak 0.0013889 of a second before the first wave started?

 

[wwhitney]

How can the results of two wave peaks that happened between time 0 and .0097222 of a second create a resulting wave + peak 0.0013889 of a second before the first wave started?

Because the "result" is not summing the two functions (waves), but taking their difference. As per post 93, 94, and 95.

Cheers, Wayne

 

[bellington]

Because the "result" is not summing the two functions (waves), but taking their difference. As per post 93, 94, and 95.

Cheers, Wayne

Then something is wrong if the resultant peak positive voltage of the two happens before the two reach their peak.