Metering 120/208 Single Phase

[Mike Scarry]

How does an energy meter calculate apparent power accurately on a 120/208 volt single phase feeder? I know a form 12s meter is usually used for this application with 2 cts and a fifth jaw for a grounded conductor connection but I am curious how the meter performs its internal calculations. This is important because I am pricing a multi tenant job where I have to supply metering but the plans call for the units to be fed from a panel board not a meter center and I don't see how the calculations work. Thanks in advance for the help.

 

[mivey]

How does an energy meter calculate apparent power accurately on a 120/208 volt single phase feeder? I know a form 12s meter is usually used for this application with 2 cts and a fifth jaw for a grounded conductor connection but I am curious how the meter performs its internal calculations. This is important because I am pricing a multi tenant job where I have to supply metering but the plans call for the units to be fed from a panel board not a meter center and I don't see how the calculations work. Thanks in advance for the help.

There are several ways but three come to mind. Old meters used thermal metering to estimate the apparent power but it was not very accurate. Then we went to kW and kvar measurements and kVA was a vectorial calculation. Arithmatic calculations are the most accurate and are simply V*I calculations made in the electronic meters. Both vectorial and arithmatic are still being used.

FWIW accurate metering requires two stators (acts like two single-phase two-wire meters), however some meters assume balanced voltages and only meter one voltage. They can be one line-line voltage combined with two shifted currents or one line-neutral voltage with the current from the other phase shifted.

 

[Mike Scarry]

So if I were doing the calc by hand, and I know how many amps are on each "leg", how would I calculate total S? I know I can just add up the power in each load, but I want to know how the meter does its calcs. Do I add the amp vectors and multiply by 120V?

 

[mivey]

So if I were doing the calc by hand, and I know how many amps are on each "leg", how would I calculate total S? I know I can just add up the power in each load, but I want to know how the meter does its calcs. Do I add the amp vectors and multiply by 120V?

FWIW, the meter uses real-time RMS data. For the arithmetic VA, you can approximate the meter function by multiplying the scalar RMS voltage times the scalar RMS current in each phase then summing the phase VA. I would not worry about the vector VA as it is not as accurate and is more complicated anyway.

 

[mivey]

I know I can just add up the power in each load

If you are going to calculate the reactive as well, you should consider that there are legitimate revenue metering debates over vectorial vs arithmetic calculations of VA especially when it comes to three-phase services and delta vs wye configurations.

Some meters measure W and var and calculate the VA by using the equation VA² = W² + var² where the vars are directly measured by phase or time shifting the currents or voltages.

Using the arithmetic method is the most accurate calculation of VA and it also is a better way to account for harmonics and their effects on equipment. Watts is the sum of the instantaneous product of voltage and current values. Watts at a harmonic is the harmonic voltage and current product times the harmonic phase angle. In the arithmetic method, VA and W are measured and used to calculate the other values. We use the equation var² + D² = VA² - W² where the D is distortion due to harmonics.

 

[gar]

130308-0941 EST

Mike Scarry:

Consider a single phase circuit. Assume no long term energy storage such as a battery. The instantaneous power to the load is p = v*i, and may be + or -. The average real power to the load for one full cycle is Pave = the integral (summation) of v*i from 0 to 2*Pi / 2*Pi. The energy for the one cycle is simply Pave * 2 * Pi. In other words the integral summation.

In the electronic power and energy IC circuits I have looked at the above operation is performed at a rapid rate, at least hundreds of samples per cycle. Thus, the low order harmonics are automatically included, and high order harmonics are essentially ignored, and assumed insignificant. The loss of accuracy from the high order harmonics becomes part of the overall accuracy of the instrument.

If you have an N phase Y supply of any arbitrary angle between phases, meaning all angles may not all be the same phase difference, then you can assume N independent single phase sources. Measurement of the product (multiplication) of the instantaneous current in a leg and the voltage from that leg to neutral, integrated over some time period is the real energy to the load of that phase over the time of integration.

Summing the individual phases will provide the total average power or energy.

.

 

[gar]

130308-1013 EST

If two TED 1000 MTUs (measuring and transmitting unit) are used with one on one phase and the other on the other phase, then power and energy from each phase can be individually measured, and these can be summed to get the total.

It may not be obvious, but a load between the two hot legs will contribute to the power measured by the two separate TED MTUs, and when those are summed the total power of the leg to leg load becomes part of the total power from the two MTUs. Essentially half of the leg to leg power is measured by one MTU and the other half by the other MTU. Draw the vector diagrams to see this.

I do not suggest that you use the TED system for this because it is not revenue grade equipment and they have a substantial problem with non-unity power factor.

.

 

[Mike Scarry]

OK, I think I've got it now. I took my loads on each "leg" and multiplied the ij components of the voltage and current vectors and the result was very close to the result when I added the apparent power in each load together. Thanks a lot for all your help.