Reconciling V, A, iPF and kW for my air conditioner

Question

On the energy meter installed by my utility company (a 1-year old digital Landis+Gyr), I noticed the following steady-state readings when my inverter air conditioner (with a variable speed compressor) was running and I am trying to figure out what could be happening Voltage: 230V Current: 2A Power Factor: 1.0 Power: 315W. Clearly the power reading is [happily] much less than expected - ordinarily I would expect power to be V times A times PF but that is not the case. To ensure the energy meter was not defective (it's just a year old) I connected a 200W incandescent lamp and got 0.75A with 175W, which shows that the meter is correct to within a small margin of error. Can I conclude that the power shown by the energy meter is lower than VxAxPF for the AC is because the compressor is probably drawing pulses of current rather than a continuous 2Amps? (I am not conversant with the exact speed control mechanism of inverter ACs.) If so would it then imply that energy meters such as the Landis+Gyr are NOT designed to show true RMS currents?

Thanks - Ram

Answer 1

No you can't assume that. A compressor is primarily a motor which will have a relatively steady current draw. The inverter may convert that to pulses, but a PFC inverter with power-factor correction would not.

But a motor has friction and requires lubrication. As the motor ages, friction will increase, and the viscosity of the lubricant changes with age and temperature. Also the load on the motor will change according to the work being done by the compressor - which is a function of the temperature on both sides.

The motor's ratings must take these factors and other variables into account. On another day, with different temperatures and a much older motor, you may find it's power is close to its rating.

All you can conclude is that today's measurements show a lower than rated power consumption.

If you need to test the power meter for accuracy, you'll need to test it with calibrated loads - both resistive, and non-linear, (such as a bridge rectifier, reservoir cap, and resistive load) not random pieces of equipment.

Answer 2

I believe that you are saying that "Voltage: 230V Current: 2A Power Factor: 1.0 Power: 315W" are all readings of the meter, not AC nameplate ratings. The inconsistency certainly indicates some difficulty with the meter. I would suspect that the AC current is distorted and the meter is displaying total RMS current rather than the fundamental and the displacement power factor rather than the total power factor. The input current distortion is due to the rectification of the input current rather than the pulsed nature of the inverter output.

The real power would be calculated as fundamental current X voltage X displacement power factor. I would expect that would be what you would be paying for unless you are penalized for distorting the current. The old fashioned rotating disk kilowatt-hour meters very effectively ignore the harmonic distortion effects and measure real power.

Can you get an explanation from the utility company?

Answer 3

I believe the following references almost completely answer my question. Thanks to Charles for helping me get there.

Reference 1: Paper titled Harmonics and how they affect power factor and Reference 2: Q&A on EnergyExperts.org

Reference 1 points out that real power equals VxAxRealPowerFactor and has a very illuminating formula and graph showing how real power factor drops with increasing distortion even though displacement power factor can stay close to unity. Also points out that "Single-phase power electronic loads such as desktop computers and home entertainment equipment tend to have high current distortions, near 100%. Therefore, their true power factors are generally less than 0.707, even though their displacement power factors are near unity."

Reference 2 clearly says "All AC electric meters account for power factor when they register and record real power. Most utility revenue meters only recognize displacement power factor."

Therefore what is most likely happening with my case is that the power factor being reported by my energy meter is displacement power factor which is 1, but the real power factor is closer to 0.7 (based on the VA product and the reported Watts). According to the formulas and figure in Reference 1, it follows that if the energy meter is correct about the real power and the RMS current, the THD of the current drawn by the AC would be about 100%.

For the benefit of those who do not want to get into the math, here is a qualitative explanation of why this happens: a [hypothetical] capacitive or inductive load may generate zero current distortion but shifts the current waveform. In contrast however, some appliances can severely distort the current but yet have the current waveform remain in phase with the voltage (i.e., the fundamental remains in phase with the voltage). That means that there is no phase shift per se so the energy meter reports PF=1. However harmonic currents which are at a frequency different from the voltage do not contribute to power, hence the total/true/real power factor is less than unity.