Power Factor and Off-Grid Inverters

[lownutHeib0]

How do off-grid pure sine inverters typically handle non-resistive (PF<1.0) loads? In normal circumstances it's assumed that the grid gives that extra current "for free" but in the much more finite world of an inverter what happens?

Would it be a reasonable rule of thumb that if no VA rating is given, the VA rating is the same as the wattage rating on an inverter?

I've been experimenting with various loads: A HPS which pulls 75W but I measured a 770VA load. My SK1500 gave an orange signal with that load, struggled harder with that than a 500-900W window air conditioner. I don't find much power factor information on that inverter. It seems to drain the batteries fast as well. I'm wondering if that extra amperage is causing the SK1500 to pump out extra heat?

Might it be reasonable for me to start calculations to install PF caps on the nastier loads?

[Mike90250]

Yes, the inverter takes the brunt of the extra work supplying the PF. But I seem to recall, it does not hit the batteries too hard, maybe someone else can follow up on that. And PF & starting surges are what my big inverter are about, if one fridge is on, and the well pump comes on, the PF & surge has to be dealt with. The inverter logs 1Kw consumption for my 1/2hp pump, that PF and efficiency sure add up and smack ya !

[Solar Sign]

From my understanding if running a ballast you need an inverter sized 3 times the load. So if running a 250 watt metal halide you need a 750 watt inverter minimum. Modified or pure sine. Not sure on the exact PF details hopefully someone can chime in.

[inetdog]

The reactive power associated with a load (VA times sin(theta), where theta is the phase offset angle) will not require any power from the batteries if the inverter is well designed. At some points in the waveform the inverter will be supplying more power than would be dissipated in the load (VA times cos(theta)) and at other points in the waveform it will be supplying less.
If the reactive portion is large enough that at some points in the waveform the net current will be flowing in the opposite direction to the voltage, either the inverter output circuit will feed the corresponding power back into the internal energy storage system or the battery or there will be major problem with the electronics.
There is often a maximum percentage of reactive power that an inverter can handle without damage, and that will be expressed as the minimum power factor (PF) that the inverter can support, or alternately the maximum VA that the inverter can deliver while the actual power output in watts remains at or below the design limit of the inverter.

As long as you operate within the design parameters of the inverter you should only be taking the real power in watts divided by the inverter efficiency from the batteries. But the inverter efficiency may be lower for a highly reactive load like a less than fully loaded motor or a badly designed switching power supply or light ballast.