Metering a Single Phase 208 VAC Circuit

[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.
---
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.

 

[aemla]

Thank you for the detailed explanation. I believe I understand now.
What prompted the question was that I knew that the meter was giving incorrect total apparent power value. So I assumed active power would also be incorrect without understanding how the value is actually obtained.

Per the explanation above, total active power is in fact the correct value.
Thank you again for sticking with me on this one.

 

[aemla]

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

One additional follow-up: does this mean that with the exact the same setup, removing one of the CTs will result in incorrect total active power measurement?

 

[Averagesupernova]

Think as if you were the watt-hour meter in question. What would you see and based on it what conclusion would you draw?

 

[Windadct]

Again - this question can be worked out with a vector diagram.

Earlier I noted that I was pretty sure your dataset (Post #32) indicated you had a Phase angle of "-105" this is telling me that you have a CT connected backwards. Did you check this ?

Edit - extra credit - which of these would a reversed CT affect and which ones would it not : Apparent, Real and Reactive Power?

 

[aemla]

Hopefully I get this right:
1. Removing one CT.
Second CT is required since addition of the scalar values is needed to get the correct answer.

2. Reversing one CT.
Phase for that CT will be rotated 180 degrees. This will affect active and reactive power (vector quantities) but not apparent power (scalar quantity).

I took another screenshot with corrected wiring:

To calculate total apparent power, Pythagorean Theorem was used with total active and reactive powers. The result would be about 501 VA.

I would think another way to solve would be via Watt's law (P=IV) where I is one of the currents and V is UL1-L2. The results would be about 868 VA.

Don't quite understand why one of these would be incorrect.

 

[Averagesupernova]

The apparent power in the load is quite simply the voltage across it multiplied by the current through it.

Never forget that a voltage across a device multiplied by the current through the device will get you apparent power without fail.

 

[aemla]

How come using Pythagorean Theorem gives a wrong answer?

 

[aemla]

Hmm, seems that previous wiring was incorrect as well.
Changed it and took another set of readings.

Performed calculations for each scenario:

First scenario appears to be the correct one.
Ran a motor with bad bearings with no load, PF seems to be reasonable.

Final phase angle also makes sense since per the image below.

I'm sure some meters with L-L configuration (few previously mentioned) could provide correct results with just one CT. However, I'm glad that correct value can be obtained through 3 phase configuration and 2 CTs.

Thank you again for all your help.