Single Phase Inverters on 208 3 Phase

bellington said:
Apparently, a single phase inverter also sees the resultant and outputs the sum of the two phases. Going back into leg A and B of 208/120Y transformer, does it sync to phase A, B, or 60 degrees between the two?
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We've already told you multiple times it's going to sync to the voltage from A to B.

So if you define N-A to be at 0 degrees phase shift (take it as the reference), and N-B to be at 120 degrees, then A-B will be at 150 degrees.

bellington said:
I want to make sure I can get power back to the battery in another building.
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You can, you don't have to worry about these details.

Cheers, Wayne
 
bellington said:
Agreed from the load side. You can always measure the results of the two waves and they can be easily obscured with many measuring devices. An oven doesn't care if it is being fed two phases offset by 120 degrees; it only sees the results of the addition of the two sine waves and the loss of 32 volts. RMS sees the result of those two waves and the loss of 32 volts. Apparently, a single phase inverter also sees the resultant and outputs the sum of the two phases. Going back into leg A and B of 208/120Y transformer, does it sync to phase A, B, or 60 degrees between the two? I want to make sure I can get power back to the battery in another building.
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Suppose on your entire grid there were one single phase 208 L-L connected load. How would that get *from* the battery? It will get back to the battery the same way. Presuming a bit about your transformers, it will be a 480V line-line source, with no current flowing on the neutral. When it gets to the battery, the two inverter phases of the battery that it's connected to will each charge at 277V, being connected to each other at the neutral point of the source. Their interaction 120deg apart will result in the single sine wave at their line terminals.
 
bellington said:
I want to make sure I can get power back to the battery in another building.
Click to expand...
What I meant by my "don't worry about the details comment" was that we don't need to be able to spell out all the details in order to be sure that the above will work. We can rely on superposition. In other words, consider the following as a thought experiment, although it's something you could actually do if desired:

1) Consider the entire premises wiring system, including the utility as a source, and the current flowing in each conductor. To write this down you'd need to make a choice of 0 degrees phase reference, and a positive direction of current for each conductor. Then you can describe the currents by attaching to each conductor a value in amps and a phase shift.

2) Look at the case where the only load that is active on this wiring system is the load in the other building, and no on-site generation. You can compute or measure the currents in each conductor. Note the real power used, this will all flow from the utility

3) Suppose your 2-wire inverter is capable of generating that amount of real power. Mentally replace that inverter with a resistive load that would draw that same amount of real power. Suppose that is the only load, and the utility is the only source. Compute or measure the currents in each conductor.

4) Now take the currents in (3) and subtract them (as vectors, since they have phase shifts; this is normal arithmetic if the phase difference is 0) from the currents in (2). This will give you the currents showing how power gets from the inverter to the load. Note that the currents from the utility may be non-zero, e.g. if you have a 480D : 208Y/120V transformer, and a single phase 208V L-L inverter, and the load is 480V L-L single phase. But the currents from the utility will work out to a net zero real power flow; the utility is just providing the necessary reactive power.

The above is all to first order and ignores resistive losses in the wiring, etc.

Cheers, Wayne
 
jaggedben said:
Suppose on your entire grid there were one single phase 208 L-L connected load. How would that get *from* the battery? It will get back to the battery the same way. Presuming a bit about your transformers, it will be a 480V line-line source, with no current flowing on the neutral. When it gets to the battery, the two inverter phases of the battery that it's connected to will each charge at 277V, being connected to each other at the neutral point of the source. Their interaction 120deg apart will result in the single sine wave at their line t
Click to expand...

wwhitney said:
What I meant by my "don't worry about the details comment" was that we don't need to be able to spell out all the details in order to be sure that the above will work. We can rely on superposition. In other words, consider the following as a thought experiment, although it's something you could actually do if desired:

1) Consider the entire premises wiring system, including the utility as a source, and the current flowing in each conductor. To write this down you'd need to make a choice of 0 degrees phase reference, and a positive direction of current for each conductor. Then you can describe the currents by attaching to each conductor a value in amps and a phase shift.

2) Look at the case where the only load that is active on this wiring system is the load in the other building, and no on-site generation. You can compute or measure the currents in each conductor. Note the real power used, this will all flow from the utility

3) Suppose your 2-wire inverter is capable of generating that amount of real power. Mentally replace that inverter with a resistive load that would draw that same amount of real power. Suppose that is the only load, and the utility is the only source. Compute or measure the currents in each conductor.

4) Now take the currents in (3) and subtract them (as vectors, since they have phase shifts; this is normal arithmetic if the phase difference is 0) from the currents in (2). This will give you the currents showing how power gets from the inverter to the load. Note that the currents from the utility may be non-zero, e.g. if you have a 480D : 208Y/120V transformer, and a single phase 208V L-L inverter, and the load is 480V L-L single phase. But the currents from the utility will work out to a net zero real power flow; the utility is just providing the necessary reactive power.

The above is all to first order and ignores resistive losses in the wiring, etc.

Cheers, Wayne
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Sorry, I should have said that I want it to get back to the battery in the MOST efficient manner possible. A balanced set of single phase 240 volt inverters would be most efficiently connected to each leg of 240 delta, or each leg of 416/240Y, to 480/277Y primary. Treating any two legs of three phase as single phase always results in losses. First impression says that is increased when back feeding. Nobody yet has demonstrated why that is not so. Now, after lots of discussion, I'm more worried and less convinced.
 
bellington said:
Sorry, I should have said that I want it to get back to the battery in the MOST efficient manner possible. A balanced set of single phase 240 volt inverters would be most efficiently connected to each leg of 240 delta, or each leg of 416/240Y, to 480/277Y primary.
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A balanced set of single phase 208V inverters can be connected just as efficiently to each leg of 208Y/120V, to 480Y/277V primary. Which transformers you already have, I gather. No need to install a special transformer to use your inverters at 240V instead of at 208V.

The only downside to the 208V configuration over the 240V configuration is that for your 180 kW of PV mentioned, if you want an AC/DC ratio of around 1.12, with 240V you'd need 7 balanced sets of 32A inverters (21 total inverters); with 208V you'd need 8 balanced sets of 32A inverters (24 total inverters). Along with the additional wiring and feeder capacity the additional 3 inverters entail.

That's a small price to pay to avoid adding a transformer.

bellington said:
Treating any two legs of three phase as single phase always results in losses.
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No, generally not, and certainly not when installed in balanced sets of 3.

Cheers, Wayne

P.S. If I understand correctly that you already have 208Y/120V distribution in the buildings, and significant loads connected to that system, then using a special transformer to run your inverters at 240V would be less efficient overall. As the 208V inverters can offset load directly on the 208Y/120V system without going through any transformer. While for the 240V solution, the power would have to pass through 2 transformers and the 480V distribution system before offsetting loads on your 208Y/120V system.
 
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bellington said:
...Treating any two legs of three phase as single phase always results in losses.
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You've had two electrical engineers and a mathematician explaining to you why that basically isn't true.

First impression says that is increased when back feeding. Nobody yet has demonstrated why that is not so....
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Rather, you haven't demonstrated why it is so. You've had several professionals try to explain your misconception to you. I'm not quite sure where we need to boil it down to in order understand where you're going wrong. Maybe Ohms Law? Getting lower power out of lower voltage does not mean you have greater losses.

bellington said:
Sorry, I should have said that I want it to get back to the battery in the MOST efficient manner possible.
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Perhaps with an exhaustive study that's beyond the scope of this forum, and included many details you haven't told us yet (conductor sizes, distances, transformer kVAs, how balanced and variable the load is, whetherthe inverters need a neutral...) one could come to an answer as to whether something like a 416/240 wye would function with less energy loss in conductors and transformers. But it seems like first we have to convince you that the difference has nothing to do with the ratio of 208/240. Well, it doesn't, and we've done our best to explain why.
 
bellington said:
Agreed from the load side. You can always measure the results of the two waves and they can be easily obscured with many measuring devices. An oven doesn't care if it is being fed two phases offset by 120 degrees; it only sees the results of the addition of the two sine waves and the loss of 32 volts. RMS sees the result of those two waves and the loss of 32 volts. Apparently, a single phase inverter also sees the resultant and outputs the sum of the two phases. Going back into leg A and B of 208/120Y transformer, does it sync to phase A, B, or 60 degrees between the two? I want to make sure I can get power back to the battery in another building.
Click to expand...
There is no loss of voltage. Yes there is a difference, but it is not technically a loss.
AC voltages are always added vectorially never arithmetically.
 
bellington said:
If connection A-B in a 3-phase wye system were actually a single sine wave (while several measuring devices would indeed average the two into one, there are two distinct waves), then A-N would be 120V, B-N would be 120V, and A-B would be 240V. But, since A-B is a combination of 120V A-N phase A and 120V B-N phase B, then the combination of those separate phases combined is the result of the addition of those two waves separated by 120 degrees and equal 208 volts, not 240. The 32 volts is lost in the combination of the two waves, with sections of A sine wave rising correspond to B sine wave falling and thus cancel each other.
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It's really a subtraction of the two voltage waveforms, since it is voltage at A, minus voltage at B (or vice-versa, a half a cycle later). The voltage isn't really lost, it's just that you are subtracting two vectors representing the waveforms, that are 120 degrees apart, instead of 180 degrees apart. Draw a triangle with two 12 cm sides, and a 120 degree angle. The remaining side will be 20.8 cm. At this scale, 1 cm represents 10 Volts.

There isn't any extra power dissipated in anything on the AC side as heat due to this, whether it is in the inverter, the AC circuit, or a transformer. The inverter is simply limited to its maximum output current, and the product of its output current and the AC voltage its electrical environment provides, is what becomes its output power. The inverter follows the AC voltage that the grid gives, and if the grid gives 208V instead of 240V, it simply produces power consistent with 208V instead of 240V if it is programmed to do so. This is usually a firmware setting to change the internal relay setpoints that tell the inverter when the grid is "healthy", and allow it to operate.

What ends up happening if you attempt to put "too much" DC power on this inverter, is that the inverter will shift its input voltage away from the ideal spot on the IV curve of the array, so that it moves it closer to open circuit voltage. The excess power remains on the modules in the form of heat, but never any more heat than the modules would generate when the system is turned off anyway.
 
If I connect two 1200 W hair dryers, one to each 120 volt leg of 208, I consume 2400 watt at 10 amps. If I run that same 10 amps through both legs connected together I consume only 2080 watts. Is the primary side of the transformer using less current when 10 amps run through two legs individually than it does running two legs together?
 
bellington said:
If I connect two 1200 W hair dryers, one to each 120 volt leg of 208, I consume 2400 watt at 10 amps. If I run that same 10 amps through both legs connected together I consume only 2080 watts. Is the primary side of the transformer using less current when 10 amps run through two legs individually than it does running two legs together?
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Sorry, is the primary side using less current when running 10 amps through both legs connected to the same load as it does when running 10 amps through the two legs individually?
 
bellington said:
If I connect two 1200 W hair dryers, one to each 120 volt leg of 208, I consume 2400 watt at 10 amps. If I run that same 10 amps through both legs connected together I consume only 2080 watts. Is the primary side of the transformer using less current when 10 amps run through two legs individually than it does running two legs together?
Click to expand...

To analyze this you need to calculate the vector current flowing in each secondary coil, figure out the vector current flowing in each primary coil, and then add things up.

I will take a very common setup for the above, with a 480V delta to 208/120V wye transformer. The turns ratio of the coils is 480:120, where a L-L coil on the primary side feeds a L-N coil on the secondary side. I'm ignoring magnetizing current, and assuming a perfect transformer.

Because the primary is delta, current on 2 secondary terminals results in supply current on 3 primary terminals.

In your first example, with 2 10A 120V loads, the primary current is 2.5A in two of the coils, with terminal current of 2.5A, 4.33A, and 2.5A, with everything in phase and 2400W being delivered to the transformer on the primary side.

In your second example with a single 10A 208V load, the primary coil current is still 2.5A in two of the coils. I'm pretty sure the terminal currents are now 2.5A, 5A, and 2.5A, but I don't have time to go into the detailed analysis. If I am correct then in the second example the primary terminal current actually _increases_.

The point I am trying to get across is that the phase angle of the current flowing in the two examples is _different_, and in the second case you will now have a power factor, with higher VA but lower true W flowing into the transformer primary to the load.

In any case, this doesn't model the situation of your PV array, because you have a _balanced_ system. A better example is if you connect three 1200W 120V resistance heaters to each of the three legs at 208V, and compare this to having three 1200W 5.77A 208V resistance heaters connected across the three leg pairs. Both the secondary and primary currents are exactly the same in the two cases.

-Jon
 
bellington said:
If I connect two 1200 W hair dryers, one to each 120 volt leg of 208, I consume 2400 watt at 10 amps. If I run that same 10 amps through both legs connected together I consume only 2080 watts. Is the primary side of the transformer using less current when 10 amps run through two legs individually than it does running two legs together?
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Note that in your first example, you are using 3 wires (A, B, N), and each wire is carrying 10A (the way the A-N neutral current and the B-N neutral current add in 3 phase is that they only partially cancel, and you get 10 + 10 = 10; see vector math). While in your second example you are using just 2 wires (A and B) and each one is carrying 10A. So in terms of "watts delivered/wire" with a fixed amps on each wire, the second case wins: 2080/2 = 1040, while 2400/3 = 800.

Now to answer your question, if you have a 480D : 208Y/120V transformer, in the first example you will get 2.5A on two of the primary conductors, and 2.5A * sqrt(3) = 4.3A on the other. While in the second example, you will have get 2.5A on two of the primary conductors, and 2.5A * 2 = 5.0A on the other. So the primary currents are slightly higher, despite less power being delivered. But the power "consumed" on the primary side is still the same as the power delivered, or 2080W in the second case. The higher apparent power due to the higher current is reactive power.

Or what winnie just said.

And just to emphasize the last paragraph of his post, your question does not reflect your application, as you aren't installing just one inverter. Rather, as you are going to install inverters in balanced groups of three, the proper comparison is between (3) 1200W 120V resistive loads balanced A-N, B-N, and C-N, and (3) 1200W 208V resistive loads balanced A-B, B-C, and C-A. Both cases will give you identical primary and secondary currents.

Cheers, Wayne
 
winnie said:
In your second example with a single 10A 208V load, the primary coil current is still 2.5A in two of the coils. I'm pretty sure the terminal currents are now 2.5A, 5A, and 2.5A, but I don't have time to go into the detailed analysis.
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Quick note, you can see that this must be true because the load is 2-wire, so there is only one current phase present. That means all currents on both the primary and secondary are in phase. So when the two 2.5A currents add in one of the primary conductors, this time they sum to 5.0A.

Cheers, Wayne
 
So to sum up, everybody agrees with 2400 watts delivered on the primary side on two legs, 2400 consumed on secondary side on two single legs added together, and only 2080 consumed on the secondary side when using two legs.
 
wwhitney said:
With only 2080 watts delivered on the primary side in that case.

No first order difference in efficiency.

Cheers, Wayne
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Two of you said the primary current would even INCREASE when the load was applied to the two secondary legs (208 V). How does primary current increase and power decrease?
 
It bears repeating that the details of this unbalanced case are not particularly relevant to the OP's application. If you take any starting configuration of 2-wire power factor 1 loads or sources, and you balance it by taking 3 copies and shifting each copy by one phase rotation, you end up with a balanced configuration. In this balanced configuration the 3 phase power factor will be 1 on both the primary and secondary sides of the transformer.

So any concerns the OP may have, justified or not in the unbalanced case, are taken care of by just installing these single phase inverters in balanced sets of 3.

Cheers, Wayne
 
(Frankly I don't think we even know how important the balancing is. Seems to me that theoretically the primary source could have plenty of VA to handle the actual level of unbalanced load and generation. Or not.)
 
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