The power factor of a circuit is the ratio of real or useful power P (something we get out of circuit - heat, light, mechanical) to apparent power S. It is a kind of efficiency. The closer the pf is to 1 the better.
$$pf = \frac PS = cos\ \theta $$
Most loads are inductive in nature. They have a lagging power factor. This means that the source has to provide more current than is needed to drive the load (real power P).
If you look at the blue triangle. Source must provide S1. Q1 (\$ Q_L \$) is lagging inductive reactive power in VARs. I lags \$ V_S \$ by \$ \phi_1 \$.
Power factor correction adds a capacitor in parallel with the load. This adds leading reative power to circuit. The Qc. P and Q1 still exist.
The capacitor acts like a source decreasing the reactive power the source must provide (Q2 = Q1 - Qc). This means S2 will be less than S1, so the current required will be less. I lags \$ V_S \$ by \$ \phi_2 \$.
The power factor has improved. Best case scenario is when \$ Q_L = Q_C \$. The triangle disappears and pf = 1.
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.
Here is a link to the study mentioned above.
Best Answer
It is the minimum power factor for full-load operation. It can probably handle a lower power factor with reduced load.
It is better in terms of reducing the losses in the generator, but it won't let you deliver more than rated watts.
Correcting to 0.8 would be good. Correcting higher might not provide much benefit.
A wound field synchronous generator will supply reactive current up to the capability of its excitation control system. That isn't quite the same thing as correcting the power factor and it doesn't involve an actual capacitor, but it does make the generator behave like it has an internal capacitor.
More about generator ratings
For an engine-generator set, the KVA rating is basically determined by the maximum current. The KW rating is determined by the power rating of the engine. A generator set rated 5 KVA, 0.8 pf, the engine would be capable of generating 4 KW. Operating at 5 KVA and 4 KW would be operating at the maximum capability of both the engine and generator. If the power is 4 KW but the pf is less than 0.8, the KVA would exceed the generator rating. If the generator is operating at 5 KVA and the pf is higher than 0.8, the KW would exceed the engine rating.
No matter how the relative ratings of the generator and engine are selected, there is only one pf at which the capabilities match. Somehow it has been determined that selecting 0.8 as the standard rating satisfies most customers in terms of buying neither an engine nor a generator that is larger than required.