# question

## Inverter switching power consumption

Hi all

We have a Multiplus Compact 12|1600|70 on a boat. How do I calculate current drawn from 12v DC to power AC appliances?

I am planning a new water maker and considering the Rainman 230V 1300W AC pressure pump. Naturally 1300W at 230V would draw:

1300W ÷ 230V = 5,65A. (P = V x I)

Would that mean?:

1300W ÷ 12V = 108,3A

Obviously switching 12V DC to 230 V AC would take its share if power on top of that, which is where I need help (assuming I'm not already wrong elsewhere)

I am planning to use a generator for the Rainman 230V 1300W AC pressure pump. But it would be nice to run the pump through the inverter off 12V for 30min or so at 12:00 on a sunny day with the solar panels helping out (solar panels: 2 x 320W parallel with Bluesolar MPPT 100|50).

Therefore, what would the current draw be on the 12V Batt bank (600Ah), powering a 230V 1300W AC pressure pump, using the Multiplus Compact 12|1600|70?

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That inverter is 93% efficient, so you need to consider that amount of loss. At 1300W is equates to about 1397W at the battery.

Another thing is that 1300W is the maximum continuous rating, you might get it to ran your 1300W pump for 30 minutes, but that depends upon the power factor of the pump. You'll be running on the upper limit of the inverter.

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Hi Christo Your calcs are pretty much on-the-money.

Let's say your pv can provide 500W. (No clouds & leaving aside inverter losses). The draw from your batts will be 1300W-500W = 800W. That's 66A, and while that's not overly excessive for those batts, you'll likely experience a substantial V sag. If you & the kit can live with that?..

The bigger issue you may face is that 1300W is roughly the rated capacity of your Multi, and if your Rainman demands a much higher start current, then that could be a gamekiller. A workaround for that could be to start it on the genny (wired through the Multi), then stop the genny.

James Bond seems to get by 'living on the edge', but not all of us are so lucky. But you could try. And if you do, good luck. :)

Edit/: Posted before noting WKirby's answer.

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Thanks @JohnC

See my question to WKirby

What max W or A can my Multiplus Compact 12|1600|70 supply in AC 230V?

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Thanks @WKirby

For clarity, what does the 12|1600|70 of the Multiplus Compact mean? I assumed 12V | 1600VA but | 70?

Am I right to say that VA is close to W but without the Power Factor? And what is the PF of my Multiplus Compact?

Since there is a 16A breaker built into the Multiplus Compact, I always assumed that it can handle anything less than 16A on the AC side. Seems like I need a bit of an education.

In essence, what is the max current that my inverter can handle in AC?

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An explanation of VA vs W may be best found on google by more qualified than I. I tend to look upon VA as the 'oomph' behind the W. Explain?, naah too hard. :)

But I've found Victron specs to be never exaggerated, and I hope they keep doing that. Like, you may get more than what you read, but not less.

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Yes, the 1600 means 1600VA, which is close to Watts if power factor is 1.0.

It's the Power Factor of the loads that you need to take note of. Incandescent light bulbs, heaters and other purely resistive loads have a Power Factor close to 1.0. Electric motors and inductive or capacitive loads (like phone chargers and LED lamps if not Power Factor corrected) have a Power Factor of less than 1.0. The Power Factor of an electric motor can change depending upon the mechanical load upon it.

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Thanks @WKirby and @JohnC

Looking at the Multiplus Compact data sheet for the 12|1600|70:

Cont. power at 25°C is 1300W

Peak power is 3000W

Then, looking at the Rainman pressure pump specs, they call it a 1300W pump but their spec for it is actually 1250W.

Therefore, I think that my Multiplus would do just fine, even if the pressure pump had a higher starting current draw. Especially since I would have a generator to power the water maker if and when necessary.

As a last side note; what would happen to the Multiplus if I overloaded it? Hoping and imagining that it would not go up in smoke, surely not, based on Victron reputation.. What build in protection does it have? Thermal breaker, fuse etc etc?

If I am going to potentially run it at its limit, these are things I should know and assess. In short, is it a stupid idea with disastrous results or kind of okay?

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It will shut it's self down if overloaded. It has thermal protection. When running at high loads the fans will kick in to help cool it, but if it gets too hot it will also shut down.
You shouldn't expect any smoke to come out of it. It just may be a nuisance if, when running on the fringe, it shuts down on you.

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Thanks all. Appreciated

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