Metering a Single Phase 208 VAC Circuit

[aemla]

I would have expected the results your attachment shows. If you connect a load between L1 and N that is what it is. Between L2 and N and that is also whatever it is. The power dissipated in the two separate loads add up. There is no reason to worry about phase.
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I'll throw another curve at you. In the real world example in question, there is neutral current. However, it will not be the difference in current between leg one and leg two. Phase difference will come into play here. In a previous post you said that current imbalances in 3 phase wye guarantees neutral current. This is also not necessarily true. If the loads are connected between the legs and neutral then yes, current imbalances will cause current to flow in the neutral. But if the loads are connected only between the legs, it doesn't matter. There will be no neutral current.

Right, if there were two loads between L1 and N, and L2 and N then that would make sense.
Second paragraph also makes sense. Although, seems to be a bad practice not to use a neutral on unbalanced loads, can't that ruin equipment (hence the need for neutral)?

Unfortunately, meter doesn't provide any measurements when the neutral is disconnected.

 

[Averagesupernova]

Although, seems to be a bad practice not to use a neutral on unbalanced loads, can't that ruin equipment (hence the need for neutral)?

Suppose I have three different resistive loads attached to a three phase wye system. L1 & L2, L1 & L3, and L2 & L3. Water heaters all of different wattages for example. Exactly how could we include a neutral even if it made sense to do so? Btw, this is scenario is done all the time in the real world.

 

[Tom.G]

I believe below is what you are describing.

Yes, precisely that.

Cheers,
Tom

 

[aemla]

Yes, precisely that.
View attachment 285580

Cheers,
Tom

This is how the wiring has been since day one. The results shared earlier are based on this wiring

 

[Averagesupernova]

This is nuts...

 

[Averagesupernova]

To me this thread has been very confusing. In my opinion the real world example you gave above does NOT reflect the hand drawn schematic you have drawn. On top of that, you mention to ignore L2 and the highlights. More confusion. You have implied that a neutral is involved, but the very first post of this thread implies not. I have given my 2 cents on why the neutral needs to be forgotten about and that it is possible and common to ignore it in a 208 volt wye situation. The way the hand drawn schematic exists, I would expect that if the power meter was configured correctly it would give you leg to leg voltages, not leg to neutral. The apparent power in the load is quite simply the voltage across it multiplied by the current through it. If this power disagrees with what your power meter tells you it is likely it is configured wrong. You are hooking it up one way, and telling it you have hooked it up a different way.

 

[aemla]

To me this thread has been very confusing. In my opinion the real world example you gave above does NOT reflect the hand drawn schematic you have drawn. On top of that, you mention to ignore L2 and the highlights. More confusion. You have implied that a neutral is involved, but the very first post of this thread implies not. I have given my 2 cents on why the neutral needs to be forgotten about and that it is possible and common to ignore it in a 208 volt wye situation. The way the hand drawn schematic exists, I would expect that if the power meter was configured correctly it would give you leg to leg voltages, not leg to neutral. The apparent power in the load is quite simply the voltage across it multiplied by the current through it. If this power disagrees with what your power meter tells you it is likely it is configured wrong. You are hooking it up one way, and telling it you have hooked it up a different way.

Haha, couldn't agree more. This is my first time dealing with power circuits. I'm used to control circuits.
The actual load is an EV charger which takes L1, L2, and ground. I mentioned to ignore highlights because that's the only screenshot I had available on hand. I mentioned to ignore L2 because I believed L2 CT was not installed correctly. The point of the datasheet screenshot was to show how this particular meter sums up powers.

I have found a meter that has the following configuration:

DIRIS A10

I believe this meter will work perfectly fine.

Previous meter did not have this configuration at all. The next best thing it had was the following:

I only have this second/previous meter and was wondering if L-L power can be calculated from data provided by this second configuration where only L-N power powers are given.

 

[Averagesupernova]

Trust me, I am pretty sure the previous meter had a configuration that takes care of what you need. It is obvious in the top diagram that the unit would be measuring the correct way because it only uses one current transformer. It is implied by this that L1 and L2 are the same current. So why can't you use configuration 3 phase 4 wires any load on the lower pic/previous meter? Isn't one load across two legs considered any load? The other two phases happen to have no load. I don't see how that can be invalid.

 

[aemla]

Trust me, I am pretty sure the previous meter had a configuration that takes care of what you need. It is obvious in the top diagram that the unit would be measuring the correct way because it only uses one current transformer. It is implied by this that L1 and L2 are the same current. So why can't you use configuration 3 phase 4 wires any load on the lower pic/previous meter? Isn't one load across two legs considered any load? The other two phases happen to have no load. I don't see how that can be invalid.

Would you happen to know which one? There are only couple pages of wiring diagrams to look at.

Yeah I agree, that top diagram makes more sense for the wiring of the actual load, however, the previous meter does not have such configuration.
I did try 3 phase 4 wire (as shown below) configuration, results were the same as shown in the previous screenshot.

 

[Averagesupernova]

The pic in post #44 should do it. I assume there are configuration setting you need to set to match the connections.

 

[aemla]

The pic in post #44 should do it. I assume there are configuration setting you need to set to match the connections.

I did try that, results were the same.

 

[Windadct]

I am quite certain you have a CT polarity wrong, the only way to get a -105 phase angle.

 

[Averagesupernova]

At this point I need to know more about the setup. To get any more help here you are likely going to have to get some known resistive loads and try various things.

 

[aemla]

Okay thank you, I'll check the setup in more detail when I can. What about a hypothetical situation? When I have readings with respect to N (L1-N voltage, L1 phase angle, L1 current AND L2-N voltage, L2 phase angle, L2 current), is it possible to calculate active power between L1 and L2?

 

[Averagesupernova]

Okay thank you, I'll check the setup in more detail when I can. What about a hypothetical situation? When I have readings with respect to N (L1-N voltage, L1 phase angle, L1 current AND L2-N voltage, L2 phase angle, L2 current), is it possible to calculate active power between L1 and L2?

I'm not following. You have two separate loads between 2 separate legs and the neutral. Those powers add up. Is this what you mean? Never forget that a voltage across a device multiplied by the current through the device will get you apparent power without fail.
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Edit: By device in this case I mean whatever you have as a load.

 

[aemla]

I'm not following. You have two separate loads between 2 separate legs and the neutral. Those powers add up. Is this what you mean? Never forget that a voltage across a device multiplied by the current through the device will get you apparent power without fail.
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Edit: By device in this case I mean whatever you have as a load.

Added power would be considered two phase power (or so I think). I'm looking for single phase L-L. If my question makes no sense (very possible) then I take it the answer is no.

I believe internal meter calculations heavily rely on configuration.
Given that my meter is setup for 3 phase 4 wire, I don't think I can get L-L data out of it. If I could could, meter would have it as an option. I did find few other meters that offer L-L data/configuration.

 

[Averagesupernova]

I'll say it again:

So why can't you use configuration 3 phase 4 wires any load on the lower pic/previous meter? Isn't one load across two legs considered any load? The other two phases happen to have no load. I don't see how that can be invalid.

 

[aemla]

Link below was an interesting read. Wiring configuration/setup is everything. Per the document, 3 phase 4 wire configuration calculates (not measures) voltages between phases (I'm guessing this is why instantaneous voltage between phases is not measured, hence, active power across phases is not provided). As previously suspected, active power is calculated by measuring instantaneous voltages and currents, like oscilloscopes.
https://www.testworld.com/wp-content/uploads/fundamentals-of-3-phase-power-measurements.pdf

 

[aemla]

I'll say it again:

I believe by "Any Load", document refers to unbalanced or balanced loads. It does not mean any amount of phases.

 

[anorlunda]

With instantaneous measurements of V and I, you can calculate the instantaneous P=VI. Square it, sum the squares for N samples of a whole cycle, then take the square root of sum/N. Now you have RMS real power without measuring any phase angles.

Phase measurements are noisy.

@Windadct poinligh out that the integral of instantaneous P divided by elapsed time is average P. RMS is not needed. I stand corrected.

 

[Averagesupernova]

I believe by "Any Load", document refers to unbalanced or balanced loads. It does not mean any amount of phases.

So just exactly what is your definition of a balanced load and an unbalanced load?

 

[aemla]

So just exactly what is your definition of a balanced load and an unbalanced load?

Exactly as you've mentioned before. If my wiring was single phase L-N with 3 phase 4 wire configuration/setup of the meter then it would work (as you've said other two phases would just so happened to have no loads). But my wiring is L-L single phase. With 3 phase 4 wire configuration, meter sees this wiring as 2 phases with load and 3rd phase has no load.

I should of said "It does not mean L-L measurements are supported."

 

[Averagesupernova]

But my wiring is L-L single phase. With 3 phase 4 wire configuration, meter sees this wiring as 2 phases with load and 3rd phase has no load.

I seriously doubt it. This is why I've asked you to get some various loads and try them.
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I'm not sure you grasp that 3 different loads across 3 different leg combinations can be considered 3 single phase loads. Yet you seem to be accepting that your meter will read this.

 

[aemla]

I seriously doubt it. This is why I've asked you to get some various loads and try them.
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I'm not sure you grasp that 3 different loads across 3 different leg combinations can be considered 3 single phase loads. Yet you seem to be accepting that your meter will read this.

This is what the internal wiring of the meter suggests. There are no sensing elements between phases, all of that is calculated. Unfortunately, I'm unable to test with various loads, just have the one 208VAC load.
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Sure, I understand that. In a three phase system using a three wattmeter method, you can simply add all powers together to get the total power. This is what the meter is doing. This can also be considered as 3 single phase loads (L-N loads).

If you actually have single phase L-L but configure the meter for three phase system with a three wattmeter method then you won't get correct results. I've previously showed the power calculation example.

 

[Windadct]

You do not need to sense L-L, this is where the vector diagram and understanding it is so important. The sensing elements are looking at the waveform, not just a V measurement like a DMM.

if you draw the 2 L-N vectors, the L-L voltage is fully defined, it is the vector sum.

If you measure the three L-N voltages, you can define the entire 3 Phase situation.

 

[Averagesupernova]

So I guess it's been implied, but I'll go over it again. The unit samples each line voltage with respect to the neutral. And you are saying since this is the case, that for instance in a 120/208 volt wye connection with a line to line load the unit will think it is X amperes multiplied by 120 volts twice to obtain the power. But we all know it is only 208 volt multiplied by X amperes. The part you are missing is that the current is out of phase with the voltage in each transformer winding. It leads in one winding and lags in the other. If you do the math you will find this to be the case and true power consumption will work out correctly.

 

[aemla]

So I guess it's been implied, but I'll go over it again. The unit samples each line voltage with respect to the neutral. And you are saying since this is the case, that for instance in a 120/208 volt wye connection with a line to line load the unit will think it is X amperes multiplied by 120 volts twice to obtain the power. But we all know it is only 208 volt multiplied by X amperes. The part you are missing is that the current is out of phase with the voltage in each transformer winding. It leads in one winding and lags in the other. If you do the math you will find this to be the case and true power consumption will work out correctly.

Yeah makes perfect sense. The math that you mentioned, would that need to be done with instantaneous voltages and currents?

 

[Averagesupernova]

Yeah makes perfect sense. The math that you mentioned, would that need to be done with instantaneous voltages and currents?

You should be able to do it with vectors, but it really sinks in and makes sense if you do it with sampling instantaneously. You should easily be able to do it on your favorite spreadsheet.

 

[aemla]

You do not need to sense L-L, this is where the vector diagram and understanding it is so important. The sensing elements are looking at the waveform, not just a V measurement like a DMM.

if you draw the 2 L-N vectors, the L-L voltage is fully defined, it is the vector sum.

If you measure the three L-N voltages, you can define the entire 3 Phase situation.

Also makes perfect sense. Sensing elements look at waveforms (aka instantaneous values), not like DMM (aka RMS values).

Right, L-L voltage is fully defined. So to calculate instantaneous L-L voltages, you would need instantaneous L-N voltages right?

You should be able to do it with vectors, but it really sinks in and makes sense if you do it with sampling instantaneously. You should easily be able to do it on your favorite spreadsheet.

So then the question becomes if the meter can provide vectors or instantaneous sampling.

You both mention vectors, does that mean:
V_L1(vector) = V_L1-N(rms) * Cos(Phase Angle_L1)
V_L2(vector) = V_L2-N(rms) * Cos(Phase Angle_L2)
I_L1(vector) = I_L1(rms) * Cos(Phase Angle_L1) or I_L2(vector) = I_L2(rms) * Cos(Phase Angle_L2)
then
[V_L1(vector)+V_L2(vector)] *[I_L1(vector)] = Active Power (Watts)
or
[V_L1(vector)+V_L2(vector)] *[I_L2(vector)] = Active Power (Watts)
?
Or something along those lines.

 

[Averagesupernova]

Concerning 120/208 Wye, voltages all referenced to neutral:
You want to take the cosine of 30° and multiply it by the line voltage of L1 then multiply by the current. That is the true power that L1 winding is sourcing. Do the same with L2. Do you understand why 30°?

 

[aemla]

Concerning 120/208 Wye, voltages all referenced to neutral:
You want to take the cosine of 30° and multiply it by the line voltage of L1 then multiply by the current. That is the true power that L1 winding is sourcing. Do the same with L2. Do you understand why 30°?

Windadct mentioned the reason behind 30 degrees: "when you look at Phase to Phase (L-L) voltages vs the Phase CT - there is a 30 Degree shift in the SENSING". It makes sense to use the phase shift between L-L voltage vs current.

What's "Phase CT"? Current Transformer, right?

Per my understanding and looking at the phasor diagrams, 30 degree shift is between V(L1-L2) vs V(L1-N). Is this statement not correct?

 

[Averagesupernova]

The reason I used 30° is because there is a phase shift of 30 between line to neutral and line to line voltage. The current will align with the line to line voltage phasor in a resistive load. So you have multiply the current by a voltage phasor that is parallel to the current phasor in order to come up with true power. Make sense?
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My suggestion is to still do it both ways. Do instantaneous sampling as well. Can easily be done on a spreadsheet. All the data is right there in front of you. Easy to wrap your head around.

 

[aemla]

The reason I used 30° is because there is a phase shift of 30 between line to neutral and line to line voltage.

My understanding as well.

The current will align with the line to line voltage phasor in a resistive load.

If you have a purely resistive load, I would agree. But what if your load has some inductive and capacitive reactance?

My suggestion is to still do it both ways. Do instantaneous sampling as well. Can easily be done on a spreadsheet. All the data is right there in front of you. Easy to wrap your head around.

To make sure we're on the same page. By instantaneous sampling, I mean sampling at a much higher frequency than the actual signal. That way an actual wave of the voltage and current can be plotted. Then pick an instance of time and multiply voltage and current to get true power (do this multiple times and average). That way only resistive components will be accounted for to get the true power.
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Other way to measure active power is to do the work, similar to how electromechanical meters work: rotate a disk. I'm not sure but I believe some digital meters do the same thing with the addition of ADC. This would allow meters to be much cheaper since high frequency sampling isn't required.
Does that seem right?

 

[Windadct]

When we say current will align - correct this is not considering the reactive elements, but referencing the relationship of the SENSING/Measurement. So yes - think in terms of resistive loads.

For ADC systems you are sampling ( digitizing) the waveform - and this has a sampling rate that should be many times higher than the fundamental. But this is not as high as typical audio sampling - so really not too high in frequency overall and not a significant cost adder. One thing that is needed is the V and I need to all be sampled at the same time - but still Digital metering overall costs much less than other methods today.

It seems you are beginning to see the light.

SO - if you ONLY had a single L-N system. V1-N and I1 - are measured and accounted for with no phase shift. ( Phasor diagram - the V and I phasors are aligned)

If now it is a three phase system - but there is ONLY one load connected L-N the power is the same - correct? ( the other Phase-Neutral voltages are present but have no impact in the POWER - so whether they are there or not is irrelevant to POWER)

Next - you have 2 separate loads connected L1-N and L2-N -- each phase phase has power - and the power of the system is just the SUM. ( power is a scalar).

Now - connect a load from L1 to L2 ( not to the neutral) -

The real current is V L-L ( vector) / Load.

each phase CT "sees" the same current, but it is "in" one CT and "out" the other. The meter evaluates each, so in one case the real current is shifted 30 Deg Leading (ahead) of the phase voltage and in the other it is shifted 30 deg Lagging(behind) the phase voltage- each is phase shifted from the Phase-Neutral voltage that it is measuring

This is becasue becasue the voltage applied to the LOAD is VL-L (vector) = VL1-N(Vector) + VL2-N(Vector) -- this is where the 30% comes in.

So again - this can be a single phase L-L load and a 3 phase meter will still read the POWER correctly. The presence of the third leg voltage is irrelevant to the power measurement.

 

[Averagesupernova]

If you have a purely resistive load, I would agree. But what if your load has some inductive and capacitive reactance?

The reason the hypothetical example is resistive is to keep things simple. Once you understand the resistive example, do the trigonometry for a 100% inductive or capacitive load and see what happens.