Calculating Power for Unique Split Phase Set Up

[renewablesHI]

Hello all,

My question today is related to a PV system that utilizes two 120 V inverters with a common neutral that generate sinusiods 180deg out of phase to create a split-phase 240V waveform. The two 'hot' outputs of the inverters, say L1 and L2, and the common neutral are connected across a single phase on the secondary side of a high-leg delta transformer (say A to B). Between the inverters and the xfmr is a three phase smart breaker. There are current probes (CTs) and voltage probes (PTs) on each 'phase' going to the breaker, the values of which are recorded by the smart breaker.

I need to know how to calculate the power coming out of the inverters using the per phase currents and voltages. The inverters are indeed grid-tie inverters and the primary side of the xfmr is connected to the utility. There are more elements to the system on the primary side of the xfmr, yet I am only concerned with calculating the power from the PV inverters.

Obviously, typical three-phase balanced power calculations will not be accurate in this case since one of the phases (say Phase C) is not connected and consequently reads no current by the meter. Summing individual, per phase power calulations seem to get me in the ballpark, yet I don't believe that method is accurate. Another possibility is to say the power is one of the phase current values times the voltage across that phase (P = Ia * Vab), yet I wonder about this also, though it does seem to provide a reasonable efficiency for the PV system.

As an added twist, sometimes the current values reported for Ia and Ib are up to 8A different. I realize this could be a problem with the data collection system, but it could also mean that the current flows through the neutral, which, of course, is not metered.

Does anyone know the correct way to calculate the total power flowing from this PV system or have any thoughts to add?


I appreciate the interest

 

[gar]

130131-2017 EST

renewablesHI:

I do not quite understand your circuit.

You have a L1, L2, and neutral from the PV inverters. I have to assume that you are connecting to a center tapped secondary on the high-leg three phase delta transformer. Really I do not care if it is three phase or single phase as the source, just that it is a center tapped secondary.

I have to assume the center tap is grounded (earthed).

You need two single phase meters one for L1 power, and a second one for L2 power. Add L1 and L2 powers.

Or use a combined meter, like the power company uses, where the total instantaneous voltage is used in combination with the instantaneous sum of the two currents, L1 and L2, properly phased.

Where you go from here to select components is a function of many factors that you have not provided. Some of the questions are: purpose of the measurement, accuracy, need to record, and accumulation time.

.

 

[renewablesHI]

More specifics

More specifics

gar,

It is indeed connected to "a center tapped secondary on the high-leg three phase delta transformer". Also, you may assume that the center tap is grounded.

Just thinking about the waveform here, imagine the voltage at L1 is peaking and the voltage at L2 is in a valley... shouldn't technically the current that leaves Inverter 1 (L1) and travels through the phase of the transformer return through Inverter 2 (L2) since their waveforms are 180deg out of phase? So theoretically they should be the same current which would make it erroneous to sum them, right? I guess in this instance Inverter 1, center-tapped leg and inveter 2 would be in series. Is that the proper way to think about it?

Thank you for you comments on instrumentation. Unfortunately the system is already built and no changes can be made. The meter currently installed makes per phase measurements. I have voltage, current, and phase shift values for each phase. The three voltage phases are 0, 120, 240 degrees (as expected), but the values I have for the phase shift between the currents and voltages of individual phases seems to be all over the place.

Does this help clarify?

 

[GoldDigger]

gar,

It is indeed connected to "a center tapped secondary on the high-leg three phase delta transformer". Also, you may assume that the center tap is grounded.

Just thinking about the waveform here, imagine the voltage at L1 is peaking and the voltage at L2 is in a valley... shouldn't technically the current that leaves Inverter 1 (L1) and travels through the phase of the transformer return through Inverter 2 (L2) since their waveforms are 180deg out of phase? So theoretically they should be the same current which would make it erroneous to sum them, right? I guess in this instance Inverter 1, center-tapped leg and inveter 2 would be in series. Is that the proper way to think about it?

Thank you for you comments on instrumentation. Unfortunately the system is already built and no changes can be made. The meter currently installed makes per phase measurements. I have voltage, current, and phase shift values for each phase. The three voltage phases are 0, 120, 240 degrees (as expected), but the values I have for the phase shift between the currents and voltages of individual phases seems to be all over the place.

Does this help clarify?

Since the inverter outputs are each 120 volts, then you can either sum the currents and multiply by 120 volts or take the identical current value once and multiply it by 240 volts. Same result either way.

Since this will be a "sell only" backfeed situation, is the smart switch wired with inverter as the line and the transformer as the load? If not, then the power factors will end up negative, which may be causing some problems. Is it supposed to be capable of handling that?

With no loads on two of the three phases, some of the currents may be zero, or very close to that. That might explain the strange per phase values. If the current in any phase is zero, the phase angle becomes very hard to measure, but the switch may be trying to anyway
As long as the metering circuit handles bi-directional power transfer correctly, the sum of the phase powers should be equal to the inverter output.
The inverter should be designed to operate with very low distortion power factor and a displacement power factor of close to either +1 or -1.
Any imbalance between the two inverters should not affect the power transfer to the 3-phase transformer.

 

[gar]

130131-2404 EST

renewablesHI:

Since you do have two independent inverters, and you have a neutral connected to the transformer center tap, then you need to treat the two sources separately if there is a substantial voltage magnitude difference between L1-N, and L2-N. Whether this difference can be ignored will depend upon the accuracy you need. Further you can not assume that both inverters are producing the same power at the same time. Example: cover the array that feeds one inverter so no power is generated, then that covered inverter will produce no power output and its line current will be about 0.

If the two voltages are approximately equal in magnitude, then you can measure the total L1-L2 voltage, this is what is done in a standard rotating disk kWh meter, divide this voltage by 2, get its average value, and multiply the instantaneous voltage by the correctly phased instantaneous sum of the two currents, I_L1 - I_L2 .

If the inverters are working as good current sources, then PF = approx 1.0 . In this situation of assumed equal voltages you could approximate the power as ( V_{L1 to L2} / 2 ) * ( |I_L1 | + |I_L2 | ) all RMS values, except if there is considerable waveform distortion, then the results may not be good.

Where there is waveform distortion you want to use instantaneous voltage and current multiplied together, and averaged over some short time to obtain average power for that short time. Then calculate the short time energy for each of the short time periods, and sum these to get energy over a longer time.

I believe you are implying that other energy is flowing thru the existing kWh meter, and you want to determine the power and/or energy from or to the inverters. To do this you need additional instrumentation in addition to what you presently have.

,

 

[gar]

130201-0819 EST

renewablesHI:

If you want to measure power/energy, and the voltage and current waveforms are arbitrary, then you must determine the instantaneous product of i and v, and process this instantaneous power value as needed.

For instantaneous power at any instant just use p = i*v .

For short time average power integrate p over the desired short time, and divide by the short time. Don't divide by time if you want energy for that short time.

For a longer time just use the longer time period.


If both voltage and current waveforms are sinusoidal, then the average power is P_ave = V_RMS * I_RMS * PF over the averaging time of the RMS converters. Short time energy is E_SHORT = P_SHORT * T_SHORT . Long time energy is the sum of all the short time energy values over the long time period.

.

 

[renewablesHI]

reply

reply

So this system is a little more complex than first explained. I didn't want to completely turn people off to responding. This is not a "sell only" system, but instead a "buy/sell" (there are batteries on board that dictate this). Since none of the values reported for current are negative, I'm assuming the meter does not report negative values and instead have been looking at the phase angle between the voltage and current (which is sketchy). Also, I do not have the Line-Neutral voltages available, only Line-Line.

More to the story, the transformer is connected to a larger AC bus network that has multiple sources and loads connected that maybe single phase or three phase. To clarify, only the PV/Inverter system is connected to the smart breaker/meter.

I think the root of my problem may reside in data collection though. As I meant to emphasize before, at times, the current reported for Ia is, say, 8A, yet the current Ib is 0A. I'm just not sure that this is possible, which means a data collection error. It is never 8A and 3A or 2A and 7A. They are either both zero, one zero or both within 1A of each other.

If it is possible, then I would need to sum the two currents, as you all mentioned. How would generating a current on only one half of the center-tapped phase affect the primary side? It all seems very strange.

 

[gar]

130201-2033 EST

renewablesHI:

In an AC circuit if you simply use a conventional AC ammeter you will never get a negative current reading because you have no input to the meter to provide phase information.

If you have phase information to correlate with the measurement of current, then you can define a + and - current. This does occur in power measurement. This is how you would determine the direction of power flow.

What is upstream of the center tapped transformer secondary is of no consequence relative to your measurement of power, and determination of what the inverters are doing. I do not believe you understand what I previously said about the two inverters and that they do not necessarily supply the same amount of power to the grid. If you look at the power to and from the grid at the primary side of the center tapped secondary, then you have no way to know how much power each inverter is producing. But on the center tapped side you can know each of the power values with the correct equipment.

I think the root of my problem may reside in data collection though. As I meant to emphasize before, at times, the current reported for Ia is, say, 8A, yet the current Ib is 0A. I'm just not sure that this is possible, which means a data collection error. It is never 8A and 3A or 2A and 7A. They are either both zero, one zero or both within 1A of each other.

I have previously explained how you can get zero current on one inverter, but I would not expect a large disparity between the two currents under normal conditions. You need to continuously monitor both currents vs time and provide us that data.

That you have batteries and how they interact with the system needs more description.

.

 

[mivey]

There are current probes (CTs) and voltage probes (PTs) on each 'phase' going to the breaker, the values of which are recorded by the smart breaker.

So you have three CTs and three PTs, withe the PTs all connected to the neutral.

I need to know how to calculate the power coming out of the inverters using the per phase currents and voltages.

If the inverters are the only thing beyond the breaker, you just sum the phase current values times the phase voltage times the power factor in each phase.

If there are loads on the same side of the breaker as the inverter, you will only get the net difference of inverter and load (and losses). If the load is greater than the inverter output, you will read the power through the breaker to the load (load minus inverter). If the load is less than the inverter output, you will read the power through the breaker to the utility (inverter minus load).

If there is no load, you will read the inverter output to the utility (less losses).

Obviously, typical three-phase balanced power calculations will not be accurate in this case since one of the phases (say Phase C) is not connected and consequently reads no current by the meter.

Doesn't matter.

Summing individual, per phase power calulations seem to get me in the ballpark, yet I don't believe that method is accurate.

Why not?

As an added twist, sometimes the current values reported for Ia and Ib are up to 8A different. I realize this could be a problem with the data collection system, but it could also mean that the current flows through the neutral, which, of course, is not metered.

Does not need to be as long as it is the common for the PTs (which I am assuming it is based on the information provided.

Blondel's Theory (which is proven to be be true) states that you need N wattmeters if you have N conductors. It also states that if voltages use one of the N conductors as a common, then you do not have to use the current for that common conductor.

Does anyone know the correct way to calculate the total power flowing from this PV system or have any thoughts to add?

Sum the per phase power.

 

[mivey]

Just thinking about the waveform here, imagine the voltage at L1 is peaking and the voltage at L2 is in a valley... shouldn't technically the current that leaves Inverter 1 (L1) and travels through the phase of the transformer return through Inverter 2 (L2) since their waveforms are 180deg out of phase? So theoretically they should be the same current which would make it erroneous to sum them, right?

You don't sum the currents because as gar pointed out, they may not be the same. Assuming no neutral current, the positive current for the inverter on its positive voltage wave becomes the negative current for the inverter on its negative voltage wave. Both inverters are delivering power because the signs of the currents equals the signs of the respective voltages.

When you have neutral current, the return current of one inverter is not the same as the sending current of the other. Now it becomes important to calculate the power of the the two halves separately, especially if the line-neutral voltages are unbalanced.

With balanced line-neutral voltages, you can use the line-line voltage and 1/2 of one current plus 1/2 of the negative of the other current, which is what a utility meter does (assuming balanced voltages is usually not a bad assumption with a strong source). This violates Blondel's Theory but is OK with balanced voltages.

I guess in this instance Inverter 1, center-tapped leg and inveter 2 would be in series. Is that the proper way to think about it?

Yes, one inverter becomes the return of the other (as opposed the the neutral being the return).

Thank you for you comments on instrumentation. Unfortunately the system is already built and no changes can be made. The meter currently installed makes per phase measurements. I have voltage, current, and phase shift values for each phase. The three voltage phases are 0, 120, 240 degrees (as expected), but the values I have for the phase shift between the currents and voltages of individual phases seems to be all over the place.

Does this help clarify?

Oh, I see. Then maybe you have three line-line voltages and three line currents. With that setup you will not be accurate for unbalanced line-neutral voltages.

Normal metering uses #1) the line-line voltage across the center-tap winding along with the two line currents (the two line-neutral voltages can be used for better accuracy) plus #2) the line-neutral high-leg voltage along with the high-leg line current. This would give you a high-leg voltage that is 90? out of phase with the other voltage (or voltages). You would then have 0? and 90? voltage (or 0?, 90?, and 180? voltages).

Let's assume you have the line-line voltage (and also assume the line-neutral voltages are balanced so we get an accuracte reading). Then you will need the line-line voltage across the center-tap winding (ignore the other two voltages since you do not care about the high-leg line current). You will then sum 1/2 of one line current with 1/2 of the negative of the other line current and multiply that sum times the line-line voltage. Which one is positive and which one is negative will depend on how you define the voltage polarity.

For polarity, be sure the line-line voltage has the "positive" terminal on the "positive" current line and the "negative" terminal on the "negative" current line.

 

[mivey]

Also, I do not have the Line-Neutral voltages available, only Line-Line.

Then you will only be accurate as long as the line-neutral voltages are about balanced.

To clarify, only the PV/Inverter system is connected to the smart breaker/meter.

Then we do not have to worry about the net that I mentioned before. You will read power to or from the inverter.

I think the root of my problem may reside in data collection though. As I meant to emphasize before, at times, the current reported for Ia is, say, 8A, yet the current Ib is 0A. I'm just not sure that this is possible, which means a data collection error. It is never 8A and 3A or 2A and 7A. They are either both zero, one zero or both within 1A of each other.

I don't see that as a problem (metering-wise anyway) as long as the line-neutral voltages are about balanced.

If it is possible, then I would need to sum the two currents, as you all mentioned.

Use the line-line voltage across the center-tap phase multiply it times the sum of 1/2 of one current plus 1/2 of the negative of the other current.

The primary and secondary currents are isolated. The only thing connecting them is a force. In layman's terms: the amount of force delivered determines the amount of current created since the voltages are relatively fixed by the net flux.

How would generating a current on only one half of the center-tapped phase affect the primary side?

The primary side would just see 1/2 the power (thus 1/2 the current). Each 1/2 of the center-tap is contributing to the flux seen by the primary side. The voltage ratios are going to stay the same. What changes is the current on the primary. Double the flux on the secondary (run the same current through both winding halves) and you will double the current on the primary since it will produce an equal counter-flux.

 

[mivey]

If the two voltages are approximately equal in magnitude, then you can measure the total L1-L2 voltage, this is what is done in a standard rotating disk kWh meter, divide this voltage by 2, get its average value, and multiply the instantaneous voltage by the correctly phased instantaneous sum of the two currents, I_L1 - I_L2 .

Utility meters use the full L1-L2 voltage and the lines are passed through a current coil in opposite directions and the current coil is wound with 1/2 the normal turns.

 

[GoldDigger]

Utility meters use the full L1-L2 voltage and the lines are passed through a current coil in opposite directions and the current coil is wound with 1/2 the normal turns.

Mathematically, that is 100% accurate only if the L1-N and L2-N voltages are identical, as mentioned earlier, OR the two leg currents and overall power factors are equal.

But it is apparently good enough for practical purposes.

 

[mivey]

Mathematically, that is 100% accurate only if the L1-N and L2-N voltages are identical, as mentioned earlier, OR the two leg currents and overall power factors are equal.

But it is apparently good enough for practical purposes.

Sometimes violating Blondel's Theorem is not so much a violation as a minor infraction.