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I'm using a new set-up of a 3.6 kW inverter with two 150 Ah gel batteries in series.
I don't understand why the inverter reports a much larger VA than the watts that are consumed. As I understand it, VA is the actual energy draw from the battery.
As an example, when the reported load is 180 W, the reported VA is 280. That's only around 60% efficiency. Is that normal?
Edit: Attaching some examples (lights are off on purpose, as they would draw +60W):
Case 1
Case 2
I think I understand your confusion.
You are trying to power a fridge, don't you?
The VA measure ("apparent" power) has only indirect relation (see the p.s.) to the efficiency. It is a property specific to the AC power setups (not only inverters at any rate).
At any given moment, the power transferred is a product of the current and the voltage in the circuit.
The AC voltage switches direction many times per second (depending on where you live, it is either 50 or 60 cycles per second, this is what the "Hz" measure is for). The AC current does the same, but depending on the load type, it may lag or advance in relation to the voltage. This amounts to the load returning some part of the energy back to the source in each cycle.
If the load misbehaves like this, the "real" (averaged) power transferred from the AC source to the load will be less than the "apparent" power (the product of the averaged current and the averaged voltage).
The ratio between these two "powers" is called a power factor. Good AC loads have a high power factor (near 1.00 or 100%), bad loads have less.
Examples of "good" loads in regard to power factors are incandescent bulbs, heating appliances and modern electronics.
Examples of "bad" loads are most types of electric motors, cheap LED lightbulbs and older electronics.
A sane and efficient inverter is expected to consume input power related to the "real" output power (W) and not to the "apparent" output power (VA). In your case, it could be something like 200W (allowing for ~90% inverter efficiency, normal for a modern inverter).
On the other hand, the inverter output stages need to be engineered for the "apparent" power that may be higher than the "real" power of the load.
This is why inverters have both "real" power (W) and "apparent" power (VA) ratings and this is why your inverter reports both values.
If you keep adding load to your inverter, you could overload it both by the "real" power consumed by the load and by the "apparent" power, independently from each other.
p.s. A minor (in your case) consideration could be that a load with a bad power factor somewhat lowers the inverter efficiency. But this effect is quite minor for a modern inverter and in your case maybe amounts to additional 5W at the inverter input.
p.s. #2 If you want to estimate the real efficiency of your inverter, you need an additional value - the power consumed at the inverter input. If the inverter does not report it (most of them don't), you need a device that measures the battery voltage and the battery discharge current. Devices like this do exist, but they don't add much value to setup like yours so they are rarely used.
p.s. #3 in regard to your comment about the awful efficiency, ~60% is quite low and neither to be expected from a brand new inverter nor particularly safe for it. This would mean that those missing 100W become heat at the inverter. This is a lot and the fan would run almost constantly. One would expect like 2kW load in order to get 100W loss. At least this is how a similar inverter of mine behaves (24V input, 2kW/4kW peak). It has label efficiency of 92-97% and is almost the cheapest I found.
The ratio of watts to VA is the power factor. Ideally, it should be 1.0 or 100%. But electronic appliances can often have very poor power factors. 0.6 or 60% is believable.
The problem here is in the loads connected, not the inverter.
Watts is the amount of energy going out, VA is the RMS current multiplied by RMS voltage, which may be a little more or perhaps significantly more than the power figure. In practice an inverter will have a maximum power capability which is limited by the magnetics and a maximum current capability that’s limited by the silicon (or perhaps copper). A designer would choose values that are appropriate for the expected load characteristics.
This is a mild deal for motors and other inductive loads, but a very big deal for electronic loads which can be choosy about which portion of the sinewave they want.
So it's your loads. If you doubt that, plug in a purely resistive load like a toaster/and you should see VA = W. You can also use a Kill-a-Watt energy monitor on utility power to read out the appliance's normal behavior on utility AC.
Traditional utility equipment like transformers have to deliver the whole sinewave. Generators slightly less so, and inverters muchly less so. If the load isn't drawing it in that microsecond, the inverter doesn't actually have to deliver it. We don't have a transformer core or generator stator winding that acts like a huge inductor to worry about.
So VA seems to be exactly the battery draw current * battery voltage.
VA has no relevance in the DC world. You might have a spiky choppy load, but in DC, stick a sufficiently large capacitor in front of it and it's not spiky anymore.
In my case the load is: a PC, two routers and a GPON terminal, plus a led light bulb, occasionally.
Why are you inverting? All that stuff will run cheerfully on 12 volts DC.
The routers probably have a wall wart that makes 12 volts DC, or you can get routers that do. 12 volt LEDs are really not a problem. A PC power supply, internally they are 5-12 volts and plenty of people sell PC PSUs that input 12 volts. Google kinda started it, as an energy saving measure, so they could run their cheap PC arrays directly off the UPS batteries instead of the inefficiency and stupidity of a double conversion.
Nothing annoys me more than off-grid people who rack $4000 of batteries specifically so they can keep using their 1980s piece of crap refrigerator that draws an average of 400 watts. They need to spend $400 on batteries and then $900 on a new fridge that averages 40 watts.
Conservation, specifically replacing inefficient stuff, must come before the battery budget.
Your biggest problem, beyond double conversion, is that ??? PC that was not designed in any way, shape or form for energy efficiency. I don't know if your application is off-grid or just a homebrew UPS for "enough time to close down databases and flush disk caches"... but if you want it to be a runner, your #1 way to save on batteries is an efficient PC, such as the 2018 Intel Mac Mini + BootCamp. (The last Intel Mini). Keep a small partition with MacOS just for maintenance, and run Linux or whatever on the bare metal. The Mac Mini runs natively on 12V and conversion isn't hard. And a 2018 Mac Mini is cheaper than a few hours worth of batteries.
