Victron Power Factor

I currently have an Infinisolar 3kW Plus and I can easily and reliably pull 3kW from the inverter during load shedding ie no grid assistance.

My understanding is that I have a good power factor because of good appliances in my home so my power factor should be very close to one based on the above.

I would like to install a Victron Multiplus II 48/5000

The rating of the inverter is confusing. Is it rated at 5kW if the power factor is one? Or is it actually a 4kW inverter that will down rate even further based on temperature and load?

The 4kW was calculated at a power factor 0.8 (think it is shown there somewhere on the spec sheet). So it is a 5kVA inverter. If you have 5kW of resistive load, it’ll power it. It does downrate itself when it gets too hot (apparently, I haven’t observed mine doing it), but this shouldn’t be unique to the Victron. Any piece of electronic equipment should throttle on heat (even your cellphone’s CPU does).

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It is is a lot simpler than this. Simply, inverters are usually speced for the peak load they need to carry, but they rarely carry the peak load continuously. The marketing department of most inverter makers will therefore stretch the number a little.

The Victron inverters can run continuously at 80% of their rated power. If you run them at 100%, they will slowly overheat. Slowly depends on the ambient temperature… but generally it’s about one hour. On a cold morning, it may not overheat at all.

So the 5kVA Multi is an actual 5kW inverter and it will handle the load just fine. In fact it will even handle a little more than that for short periods (ranging from a few minutes, to about 1 second for 200% Pnominal). But if you want to run a continuous load, then it’s only a 4kW.

Power factor really has nothing to do with it in this case. Inverters are rated in kVA simply because the peak current is the limiting factor rather than the real power.

Ah okay! But what I did notice is that the inverting capacity seems to be on the input rather than output? I.e. 5kW input on the DC side will be inverted and you end up with around 4.5kW AC. Or was this just my eyes playing tricks on me?

Efficiency. Only between 85% and 93% is effectively converted into AC power.

Thanks for the clarification!

my understanding is that a geyser is a resistive load, so a 3 kW geyser element should same as 3 kVa, as in, when connected to the 3 kVa Multiplus II, or am I missing something

I saw and experienced that. Used to run a 3kva heating 2 x 2kw geysers consecutively midday. 2.4kw was 100% ok for geyser heating and general continious house loads. Come summer I noticed that I started getting 1.8kw from the inverter, fans are running lekka, as the ambient temp was like +26 deg plus.

So I sold the 3kva on, got a 5kva, which derates to about 3kw on really hot days, problem solved.

Correct. But what I am getting at here is that power factor really has nothing to do with how the Multiplus is rated. It’s common to think that way, because this is a trick used by generator salesmen.

For a generator, the power of the engine is often the limiting factor. The engine cares about real power. So the base number you work from is real power, and then the salespeople can slap a bigger number on it by assuming a lower power factor.

The 80% grid-tied limit has nothing to do with an assumed power factor. It is done merely because running at 100% for prolonged periods would overheat it.

I can confirm that at -10°C, it can run at 100% indefinitely.

TTT is also right. If it gets hot (while running grid-coupled), it will derate even further (to 60%). I’ve run a 1.6kVA for years through the hot summers in a garage without air conditioning of any sort, and it never derated in that manner. I can only conclude that for things to go that far, it must be really really hot where you are.

thanks @Plonkster, i am changing to a 2 kW element, so should not be drawing from the grid, so the only high wattage load will be the tumble dryer which will draw some from the grid.