Harmonics in power calculations

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The discussion revolves around interpreting data from a 3-phase electronic meter, focusing on power calculations under different load conditions. In a loaded snapshot, the user successfully calculated watts, VA, and VAR, but in a no-load scenario, the calculations yielded significantly higher output values than expected. The user noted that harmonic VA and harmonic power factor (pf) were much larger in the no-load snapshot, suggesting they may play a crucial role in the calculations. There is uncertainty about how to incorporate these harmonic values into the power triangle calculations. The conversation emphasizes the need to understand the impact of harmonics on power measurements in varying load conditions.
Sparky_
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Greetings,

I’m looking at the output of an electronic 3-phase meter.

I have 2 “snapshots” from the meter.

In one snapshot I can make sense of most of the data presented.

The output gives me Va, Vb, Vc, Ia, Ib, Ic, and their angles, a single power factor, watts, VA, and VAR.

When I built my power triangle I almost agree with the outputted VA, watts, and VAR. I am within a little on some and right on it on other numbers.

For this snapshot – there was “load” - current in all 3 phases is around 1.1A.

The meter also provides “Harmonics VA and Harmonic pf” I did not use these 2 numbers in my power triangle.

My next snapshot was without load.

I am given the exact same data – current is around 0.1 in all 3 phases.

I cannot get any where close to the power triangle – meaning using the same procedure above – (Vrms)(Irms) – VA, VA(cos()) = watts and so on. I do not get the outputted watts, VA and VAR. The outputted powers are quite a bit higher.

One thing that stands out is that in this snapshot (no load) my harmonics VA and harmonics pf are quite a bit larger than the loaded snapshot.

With load: harmonics VA = 25.7 and harmonics pf = 0.069.

With no load: harmonics VA = 493.965 and harmonics pf = 0.30.

Here is all of the data for the no load:

Va = 113.336, Ia = 0.087: Ia lags Va by 15.86 degrees
Vb = 112.676, Ib = 0.136: Ib leads Vb by 21 degrees
Vc = 113.512, Ic = 0.158, Ic lags Vc by 10 degrees

Watts = 710.078
VArms = 1644.563
VAtd = 1568.625
VAR = 1398.719
Pf = 0.432

Harmonics VA = 493.965
Harmonics pf = 0.3

Can you show me how to get the (above) watts, VA rms, and VAR from the provided currents and voltages?

How do I use the provided harmonic data in power calculations?

Thanks
Sparky
 
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keep in mind what you have involved in the load, a sure mix of inductors and capacitors which can balance you pf. why your b phase is capacitive without load i am not sure, but it is to be expected inductive. remember your power is coming from a transformer, an inductive coil.

Sparky_ said:
Here is all of the data for the no load:

Va = 113.336, Ia = 0.087: Ia lags Va by 15.86 degrees
Vb = 112.676, Ib = 0.136: Ib leads Vb by 21 degrees
Vc = 113.512, Ic = 0.158, Ic lags Vc by 10 degrees

Watts = 710.078
VArms = 1644.563
VAtd = 1568.625
VAR = 1398.719
Pf = 0.432

is this data the data your meter provided?
 
Hey Sparkey

yes that is data straight from the meter. Again in the first snapshot - I could reproduce the Watts, VA, and VAR (which were provided from the meter) by using the provided currents and voltages and their angles.

In this snapshot the same calculation method does not work and I noticed that my harmonic VA and pf harmonics stand out when compared to the first snapshot. I did not use these 2 pieces in my first calculations and my gut says they are needed here. I just don't know how to use these two numbers nor what they are telling me.

Thanks
-Sparky
 
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