Active Power Measurement at Very Low pf (0.003)

I wouldn't. I would use those numbers and to hell with it.

The real power consumed by the coil will be turned into heat. I recommend you measure the heat produced by measuring the temperature gradient through a single, dominant thermal conductive path that the thermal conductivity is well known. Make sure to take enough time to see thermal equilibrium.

The best way to validate any measurement technique is to compare with another measurement technique.

I suspect that if you can do an error analysis of your two power meters that you will find your calorimeter measurement will reside in an overlapping uncertainty region of the two power meters.

As SimpleMind has said, measure the resistance and calculate I squared R. To measure a small resistance accurately, you need to use the 4 wire method where a test current is passed through the resistance and the dropped voltage is measured directly across the unknown resistance.

http://en.wikipedia.org/wiki/Four-terminal_sensing

Try the theory:

A 205 mH, at 60Hz, has an inductive impedance of 77.24 ohm

Assuming a voltage of 15000Vac (just a number) the current that you have measured in a load Z= √(R2+XL2) is 177A. I would find Z=84.75ohm and R = 34.9ohm.

The angle between U and I should be

φ = arctan(XL/ R) = 66°

Now you can measure the resistance and calculate directly the angle.

What you are saying is that you have a 2MVA shunt reactor rated at 11kV, 50Hz, and want to measure its losses.

Yes, you would NOT measure the losses in this reactor with any sort of ordinary wattmeter as ordinary wattmeters are very inaccurate near zero power factor. You would need a "low power factor wattmeter" that exhibits very low phase angle errors and is intended for such low power factor measurements. In addition to this, you would need a voltage transformer to step down the voltage - 11kV to 110V, and a current transformer to step the 177A to something the wattmeter can handle, . . like a 200/5 CT. Both the VT & CT need to be very accurate - the ordinary commercial variety used for metering will just not do.

Once you assemble this SPECIAL TEST EQUIPMENT, you are likely to obtain an accuracy of +/- 9% on the power reading. These errors would be attributed to phase errors of +/-1 minute in each of the components - the wattmeter, the CT, and the VT. This one can compare to an accuracy of about +/-300% in the power measurement when using ordinary CT, VT and wattmeter. Yes, in fact the wattmeter may well read negative power.

A more accurate method of measuring the losses of the reactor is to use a high-voltage bridge with a loss-free standard capacitor to scale down the voltage and a current comparator to scale down the current. You may well receive an accuracy of +/-1% in the power measurement at your power factor of 0.003.

I see several suggestions that you make assumptions about calculations using the inductor's ac resistance. The ac resistance will no doubt be considerably higher than the dc resistance. I'd approach this as a bench measurement, and resonate the coil at about 60Hz with a paralleled 33uF film capacitor. You can drive this from an ordinary bench function generator, e.g., through a 1k or 10k resistor, and monitor with a 'scope. Vary the frequency and measure the Q, to directly determine the AC resistance. Then you can accurately calculate the losses in service at 177A.

Yes, the ability to measure the real in power in these situations is difficult with normal meters. I was in a factory contact engineering position in a ftransformer factory that built some HV air-core reactors for utilities. (Cast in cement) While your question is for MV application the principal is the same. The resistance is driven as low as possible to try to eliminate in phase losses. Yes, heat loss is one way to cope with measuring losses in these situations. Instruments designed to measure vars only can be used and the in-phase losses calculated by comparing the Volts X Amps to the Vars. We did have instrumentation to read the losses directly, and very accurately, but we had to verify accuracy frequently. Some customers (utilities) wanted to make their own checks, so I remember various excersizes in reading osilloscopes, etc. I was only out of college about 5 years, so my math was still solid, thankfully. Instrumentation to measure this accurately is very expensive, and utility grade or better. It is of a nature that it is permanently installed as a custom installation. Usually it takes the form of a relay actually, which trips if the coil senses a shorted turn.

TRENCH Electric is the leader in manufacturing LARGE, air-core reactors. They us a transformer-arm capacitance bridge to measure losses. They have to mount the test reactor in the middle of the parking lot for the measurement, . . . as the losses in the steel components of the test station increase the losses substantially.

Yes, Trench is a leader now, but not then. Any steel or conductive loop in the area surrounding the reactor has an effect on it, but there are other ways of shielding them without resorting to large distances.

Have you tested the Power Cell by Load Controls Inc.?http://www.loadcontrols.com/products/products.html

At what AC voltage or AC current level are you interested in making power measurement? Check AC rms curernt and pass same rms DC current through the coil and you can use DC power measurement- why use AC at all? You will get (I^2)*R same whether it is AC or DC.

No, the current will be significantly lower with an AC voltage than an identical DC voltage. By definition a very low power factor has significantly more reactive impedance than real impedance.

With over 100 amperes of current I expect that the wire thickness may be large enough for skin effects to change the resistance, too. So a DC resistance measurement may not be an accurate enough resistance measurement at 50HZ.

There are no skin effects at 50 Hz. Hence DC measurements and AC power measurements will not be different. Since he has high inductive load and low resistance and PF is low- surely he is facing probelm with measurements.So I feel DC measurement will be the best.

Really, so you know with absolute certainty that the copper conductor used to handle 177 amperes of current is smaller than the 9 mm skin effect depth for 50 Hz.

Go to my link and do the math. The low resistance and high current implies to me a fairly thick conductor.

Quote: "There are no skin effects at 50 Hz"

You are wrong. There is definitely skin effect at 50 and 60Hz. Look at the following for values. marc's pages --Skin Effect

The last time I saw any thing printed doubting the Skin effect of 50-60 Hz was published in the 30's. Are all the Utility Companies wrong? I don't think so.

Thanks & noted. At very small currents it is negligible. But at 177 A that has been mentioned skin effect is significant even at as low as 50 or 60 Hz.

This can be measured at full power by using a modern digital scope with a current probe and a voltage probe.

1. First de-skew the two probes.

2. Calibrate the two probes and acquisition circuits to assure true V and I.

3. Trigger the scope from the zero crossing of the incident voltage.

4. Calculate the power waveform using scope's math.

5. Compute the FFT real and imaginary components.

6. Use a cursor to read off the REAL power (the in-phase component), and the IMAGINARY power (the out of phase component).