offgridQLD wrote: So is it correct to assume in a offgrid system (battery - inverter) If you run loads with a bad PF (motors, fluro lights) Then you will increase the strain on your wiring and inverter but the strain (load) on the battery side will not be affected.
Essentially, yes. For an inductive or capacitive load (inductive is far more common), the instantaneous power essentially circulates between the battery and the reactive elements in the load. Inductors can store magnetic energy, and capacitors can store electrical energy. In an inverter, or just connected to the mains, they can return energy almost as effectively as they can take it.
I happen to have been doing some of these power calculations at work yesterday. Here is a classic power triangle:
This represents the situation of a 0.8 power factor load. Let's say the mains is 250 V today; it's often close to that and it makes the maths a little easier. So we see 6 A flowing in the cables; that appears to be 6 x 250 = 1500 W, but actually there is only 900 W of real work being done. We say that the apparent power is 1500 VA (Volt Amps), and the real power is 900 W. VA have the same units as real power (Watts), but we like to distinguish them because they act so differently. The "discrepancy" is due to the 900 VAR (Volt Amperes Reactive) in the inductance of this load. I've drawn the 900 W at a right angle to the 1200 W, because these powers add according to Pythagoras's theorem (S^2 = P^2 + Q^2). Note that 1200 + 900 = 2100, which is not the same as 1500 at all. The reason for this is that the voltage across the inductance and the resistor representing the real load are at ninety degrees as well. It's a bit beyond the scope of this post, but you can think of these lines as being snapshots of vectors that rotate at the mains frequency (50 times per second).
So while there is 250 V flowing at the mains, there is effectively only .8 x 250 = 200 V that is across the resistance of the load, in phase with the mains voltage, doing real work (mechanical load, heat, and so on). There is a reactive voltage of .6 x 250 = 150 V across the equivalent inductor, representing the reactive part of the load. (Think if the load as an ideal inductor in series with an ideal resistor; they form a voltage divider.)
The acute angle in the diagram is the inverse cosine of 1200/1500 (simple geometry; my daughter does this stuff in grade 10 maths). It's about 37 degrees. Sometimes you see this angle called "phi", which is the greek letter ø or Φ (I hope that shows up OK). The cosine of this angle is the power factor, which is just the ratio of the real power to the apparent power. You might see "cos phi" or "cos ø" on the spec plate of a motor, for example. This is the electrical angle between the voltage and the current. As this angle approaches 90 degrees, the real power approaches zero (cos(90°) = 0). As this angle approaches 0 degrees, the real power approaches the apparent power (cos(0) = 1.0 or 100%).
Without that inductance, we could have delivered the 1200 W of power to the load using only 1200/250 = 4.8 A, or 80% of the actual current. So there is 100% x 6 / 4.8 = 125% as much current as we really need. There is some copper loss associated with the extra current, so this will cause extra I^R heating, in fact (1.25)^2 times as much, or about 1.56 times the copper losses. This extra loss has to be supplied by the battery, so this is the reason that in practice, the battery current will be higher than if we had a pure resistive (100% power factor) load of 4.8 A. It won't cost us 56% extra battery current, of course, since the losses are small compared with the total power. If the losses were 10%, then we'd have about 5.6% extra current from the battery.
Now this is all for inductive or capacitive loads, which operate linearly. For a non-linear load, such as when a diode rectifier is involved, none of the power ever comes back to the mains, so it gets more complicated. I've seen texts where they introduce a "distortion factor" D, with S^2 = P^2 + Q^2 + D^2. But it's by no means universally accepted. Most texts say that the reactive power component, Q, is simply undefined when the current is not sinusoidal. Unfortunately, I don't know enough about this to say what happens to the battery current if you have say a 60% power factor load due to a full wave rectifier with no smoothing on the mains side of the diodes. Perhaps others can weigh in here.