A transformer winding is generally more inductive than resistive thus there is a considerable phase difference between voltage and current, so wouldn't the supply power factor be very very poor in case of residential transformers where no means for power factor corrections are used?
Electronic – supply power factor of a residential transformer
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The point you seem to be missing is that it does not require power transfer from the device back to the power line during part of the power cycle to have less than unity power factor.
There are various ways of looking at what power factor really is, although they all come out the same mathematically. One way is the ratio of real power delivered to the product relative to the RMS voltage and current. If the current is a sine (let's consider the voltage always a sine in this case, since the power line has such low impedance), then you have unity power factor when it is in phase with the voltage, and 0 when 90 degrees out of phase. In the case of a sine, power does have to flow back to the line during part of the cycle to have less than unity power factor.
However, lots of other waveforms are possible. You can have current that is always 0 or positive when the voltage is positive, or 0 or negative when the voltage is negative, but that is not a sine. The spikes you mention caused by a full wave bridge are a good example. Power never flows back to the power line, but yet the power factor is less than 1. Do some examples and calculate the RMS current drawn by a full wave bridge. You will see that the total real power drawn from the power line is less than the RMS current times the power line voltage (again, we are assuming the power line voltage is always a sine).
Another way to think of this is that losses in the tranmission system are proportional to the square of the current. The full wave bridge draws its current in short spikes of high magnitude. Because of the squared nature of the losses, this is worse than the same average current drawn more spread out. I you work out that math, you realize that the way to minimize the average square of the current is to make the current be a sine in phase with the voltage. That is the only way to achieve unity power fator.
Yet another way of looking at this, which you alluded to, is to think of the Fourier expansion of the current. We are assuming some current waveform that repeats every power line cycle, so it has a Fourier series. Any such repeating waveform can be expressed as a sum of a series of sine waves at the power line frequency and positive integer multiples thereof. For example with 60 Hz power, the waveform is a sum of sines at 60 Hz, 120 Hz, 180 Hz, 240 Hz, etc. The only question is what the amplitude and phase shift of each of these harmonics are. It should be obvious that only the fundamental (the 60 Hz component in this example) is capable of drawing any net power from the power line, and that only to the extent it is in phase with the voltage. Since all components are sines, each will draw power during part of the cycle and return the same power at another part of the cycle, except for the in-phase component of the fundamental. So your way of looking at power factor as having to put power back during part of the cycle is valid if you break up the current waveform into sine wave components. However, it is possible to have a set of sine wave components that take and return power to the power line at different times such that the net from all components at any one time is zero or positive. The full wave bridge current is one example of such a waveform.
I found a study that showed various brands of LED bulbs rated 3 to 8 watts with power factors ranging from .48 to .79. If you can find 8 watt LED bulbs with .5 power factor that give equivalent light as your 40 watt incandescents, you would need 16 VA per bulb vs 40 for the incandescents. If you distribute the bulbs equally among the phases, you should not have any difficulty with the neutral currents. You should still have some concern about extra heating in the generator due to harmonics. It is difficult to determine how much the generator should be oversized for harmonics. The generator manufacturer may have a recommendation.
A harmonic filter would both reduce the harmonics and increase the power factor. I don't believe that you should purchase a 3-phase choke and harmonic filter separately. You should be able to get the most effective filter if it is purchased as a package.
Estimating Harmonic Distortion
For estimation purposes, it can be assumed that the power factor of the fundamental current of an LED bulb is 1.0. It can also be assumed that the source voltage is not significantly distorted. Total power factor = Watts / (Voltage X Total RMS current). Total or “true” RMS current is the RMS value of the distorted current waveform. It calculated as the square root of the sum of the squares of the fundamental current plus each of the harmonic currents. You can break that down as Irms = (If^2 + Ih^2)^.5 where If is the fundamental current and Ih is the square root of the sum of the squares of the individual harmonics.
If the voltage, total RMS current and power is known for an LED bulb, the harmonic current and total harmonic current distortion can be calculated as follows:
Only the fundamental current (If) produces power (W). W = V X If X pf. Assuming pf for the fundamental = 1, If = W / V
From Irms = (If^2 + Ih^2)^.5, Ih = (Irms^2 – If^2)^.5
Total harmonic current distortion, THDi = (Ih^2 / If^2)^.5 = Ih / If
The total RMS current and power may be marked on the bulb. If it is not marked it can be measured with an inexpensive power meter like a Kill-A-Watt.
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Best Answer
Unloaded transformers look like inductors. However, a resistive load on the secondary makes the primary look more resistive too. For any decent transformer, the primary looks mostly resistive when the secondary is loaded with a resistance near the power limit of the transformer.
So the power factor is very poor when a transformer is drawing little power, and gets better as the power increases. Fortunately, you care more about the power factor at high power levels.
All that said, the grid does generally look partially inductive. Large industrial motors generally present a more inductive power factor than that added by typical transformers.
Large electric customers that pay extra for a low power factor usually have a system of compensating for the inductive power factor. This is usually banks of capacitors that are switched across the power line to cancel out the inductance.