How many Watts @ 12v to provide 600 watts @240v Via Inverter

Watts will be the same apart from inverter loses. It is the current that will change.

600 watts at 12v is 50amps but with inverter efficiency it will be 55 amps.

600 watts at 240v is 2.5 amps.

Victron rate their inverters in VA so watts will be less unless the load has a zero power factor.
 
If you are taking 600w then Around 50 ah plus losses

Edit
Lenny beat me to it
 
You may be confused by terminology?

A Watt is a measure of units of energy transfer.

Regardless of the voltage (12V or 240V), the wattage would remain the same. The current flow, Amps, increases or decreases to reflect the Voltage, so Wattage remains the same.

Watt = Amp x Volt

But inverters have an inefficiency so although appliances wattage 12V or 240V might match, you do need to allow a bit of extra energy loss for the inverter.

Edit: Lenny was more succinct.
 
Lenny HB said:
Watts will be the same apart from inverter loses. It is the current that will change.

600 watts at 12v is 50amps but with inverter efficiency it will be 55 amps.

600 watts at 240v is 2.5 amps.

Victron rate their inverters in VA so watts will be less unless the load has a zero power factor.
Click to expand...
so Solar/amps @ 12v for 600w / 240v is?

Sorry its late and A, had a beer and B lazy
 
If you can help, please?

I require 600 watts @ 240v

From a Combination of Solar and 110amp Alternator via a 105ah Lithium and Victron inverter with engine running ?

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I think DC power = AC power / inverter efficiency
So to illustrate and keep the calculation simple we say the inverter is 100% efficient (it won't be), so DC power would be 600 watts / 12 volts = 50 amps if the inverter was 97% efficient it would be 618.5 watts / 12 = 51.5 amp draw.
Someone who's got a better memory than me can correct me if I have remembered the formula incorrectly.

Edit: you all beat me to it, I blame it on the scotch.
 
240 volts means each amp of current carries 240 watts. So 600W at 240V needs 600 / 240 = 2.5A.
12 volts means each amp of current carries 12 watts. So 600W at 12V needs 600 / 12 = 50A.

An inverter converts power at 12V to power at 240V. The power input and power output are about the same: 600W in this case. It converts power at low voltage, high amps to power at high voltage, low amps. Watts = Volts x Amps.
12V x 50A = 600W = 240V x 2.5A

In reality, the inverter is not 100% efficient, and losses can be around 5% to 10%. So to get 600W out you have to put about 660W in. That's why you'd have to put in about 55A, not 50A, to get the required 600W output, as Lenny HB said.

If you're using 55A for the inverter, that could be taken from the 105Ah lithium battery. In one hour it would use 55Ah, so there would be 105 - 55 = 50Ah left in the battery. In other words it would go down from 100% to 48% in one hour.

Sorry, not very succinct :giggle:
 
EuroTrotters said:
If you can help, please?

I require 600 watts @ 240v

From a Combination of Solar and 110amp Alternator via a 105ah Lithium and Victron inverter with engine running ?
Click to expand...

To power a 600W device from mains (240V) power from a 12V dc source (via an inverter) will require the following current from your dc source (either your solar or your alternator):

600W/12V = 50A.
However, as others have said you’d actually need about 55A to account for inverter inefficiencies.

To Deliver 50A from solar you’d likely need in excess of 600W of solar (more if not at the height of summer).

To deliver 50A from your alternator, you’d probably need a 50A+ DC to DC charger (and possibly upgraded wiring). It’s unlikely that a standard split charging system would deliver what you’re looking for.

If you were only ever intending to demand your 600W load with the engine running then the solar would provide some power and the requirements on each system would be reduced. Consequently, you could possibly manage with 25A from your alternator (some standard systems might even struggle to deliver that level of current) and 25A from your solar. HOWEVER, in REALITY, this is not really a practical (and real world) solution for your scenario.

In practice, you have a leisure bank and it is this that will power your load. Your solar/alternator need only replenish the energy that you have taken out of your battery and they can do this over a longer time period. Consequently, those do not need to be sized to power your load.

Ian
 
If you have a 100Ah leisure battery it will give you 600W @ 240V for about an hour if lead acid or 1.5hrs if lithium.
 
EuroTrotters said:
or maybe simply.

How many watts @12v provides watts @240v Via Victron Pure Sinewave Inverter please?

ET
Click to expand...
Just did a test with my 700w water heating, and multiplus 2kva. The water heating draws on AC output, exactly 695w, and out of battery, that’s including inverter, it shows a 765w DC leaving the battery. So the inv. efficiency in this case is about 90%. The efficiency It does change with different loads. Heating, resistive it’s the most efficient,
 
EuroTrotters said:
If you can help, please?

I require 600 watts @ 240v

From a Combination of Solar and 110amp Alternator via a 105ah Lithium and Victron inverter with engine running ?
Click to expand...
You will need to supply about 660w DC from battery, for a 600w AC load at the inverter side.
 
The efficiency of an inverter is primarily down to the ratio of the load to the inverter's rated output. Most are at least 90% efficient from half power up, and nearer 95% the closer you are to its continuous maximum.

More expensive ones have extra circuitry to minimise the quiescent (no load) current draw.

Dave

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DABurleigh said:
The efficiency of an inverter is primarily down to the ratio of the load to the inverter's rated output. Most are at least 90% efficient from half power up, and nearer 95% the closer you are to its continuous maximum.

More expensive ones have extra circuitry to minimise the quiescent (no load) current draw.

Dave
Click to expand...
Not on 12v, the only 95% efficiency you will ever see is a 5kva /48v at 70-80% load. And that with a load of power factor 1.
The 12v inv. highest efficiency ever built by any is 93%. At PF1.
 
EuroTrotters said:
All great advice. Thank you .

I am going to go with the solar Raul linked in my other thread

And victron charge controller from RogerIvy
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I also gonna upgrade my solar when I’m back from this trip. My panels will be 2 eurener 400w. I have been searching for weeks, and nothing comes close to that choice.
 
One thing that makes me always wonder is when people are/think of 'using stuff from solar panels'. The panels are connected to the battery (not directly, but it's not important here) and all devices are connected to the battery (also, maybe not directly). The battery enables that everything can be run even without solar at all, and also with 'too small' solar system. Use your stuff from the battery and the solar system (no matter what size) will eventually charge the battery.

Like in this case, with a 100Ah battery the original poster can use his/hers 600w/240v device via inverter for about an hour, no matter what size of solar system is install. And if one is installed, it will eventually charge the battery. Using something 'from solar panels' isn't really possible in any case as there are clouds in the sky and the panels will not provide constant power.

Or is there something I don't quite get about using solar power?
 
PDrill said:
Or is there something I don't quite get about using solar power?
Click to expand...

No, you've just said what I said, in an earlier post. 👍

bigtwin said:
In practice, you have a leisure bank and it is this that will power your load. Your solar/alternator need only replenish the energy that you have taken out of your battery and they can do this over a longer time period. Consequently, those do not need to be sized to power your load.
Click to expand...

Ian
 
I usually work on you will need 26 amps at 12 volt dc. To produce 1 amp at 240 volt a.c. Allowing for inverter and cable losses. If using lead acid remember the 50% rule or you will seriously reduce their ability to function. Even Lithium have limits on the maximum current drawn for long periods of time.

So 600 watts divided by 240 = 2.5 amps. 2.5 X 26 amps = 65 amps @ 12 volt d.c.

So if using a 100 amp lead acid battery when new and fully charged I would keep it to 20 mins max and whatever you take out has to be replaced
 
Bessy765 said:
I usually work on you will need 26 amps at 12 volt dc. To produce 1 amp at 240 volt a.c.
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I’m curious, what led you to arrive at the 26A figure? It’s significantly more conservative than the assumption used by most folks.

Most tend to calculate it as 600W/12V = 50A +10% for conversion inefficiencies giving 55A.

For simplicity, I tend to take the wattage and divide by 10 (instead of 12) as this is both a simple calculation to undertake and includes a conservative contribution for inefficiencies. In this case my result would be 60A.

Ian

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Why you beating yourselfs up and work around it the hard way? Work in wats. Amps are meaningless without volts, and volts changes with load and charge source. So you never can account for the energy/ power in amps correctly, never ever. One amp or 60, at 12.5v will not be the same power at 13.2v. A lithium with a bit of solar can sustain a 13.2-13.4v even under load. Some heavier load will take it down to 12.7-12.8v, or even 12v if near empty. All lithium batteries have the spec in wh for a reason. Forget the ah from lead, lithium is different
 
Raul said:
Why you beating yourselfs up and work around it the hard way? Work in wats. Amps are meaningless without volts, and volts changes with load and charge source. So you never can account for the energy/ power in amps correctly, never ever. One amp or 60, at 12.5v will not be the same power at 13.2v. A lithium with a bit of solar can sustain a 13.2-13.4v even under load. Some heavier load will take it down to 12.7-12.8v, or even 12v if near empty. All lithium batteries have the spec in wh for a reason. Forget the ah from lead, lithium is different
Click to expand...

TBH, I always want to use 13V but figure folks would inevitably ask why so I just stick with 12V.

However your point about Wh is a useful reminder.

Ian
 
bigtwin said:
TBH, I always want to use 13V but figure folks would inevitably ask why so I just stick with 12V.

However your point about Wh is a useful reminder.

Ian
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You shouldn’t, the nominal voltage 12.8v is to be used, IF you make use of all capacity. If you only cycle the top part 13+ then wh is the correct measurement.
Lead 12.7 full LFP 13.4
Lead nominal 12 LFP 12.8
Lead empty 10.5 LFP 10
As you can see above, voltage is important to be accounted for, as when you work in amps only, impossible to work out the power/ energy. Ditch both in favour of one, that’s always right at any given time. Any load takes power in W.
 
PDrill said:
Or is there something I don't quite get about using solar power?
Click to expand...
What you say is true. What I would add is that It's a good idea to size your system so that the total amount of solar power harvested over the whole day is enough for your daily consumption of power.

Each 100W of panel will harvest about 30Ah to 40Ah of charge over an average summer day - more if you are lucky. That's about 360Wh to 480Wh of energy. So if you are using say 60Ah of charge per day (=720Wh) then you probably need 200W of solar panel to keep up with your power demand. If you only have 100W of panel, then the battery will gradually empty as you take out more than you put in, so you'll have to do something after two or three days - plug in a hookup, go for a drive or fire up a generator.
 
A few posts on here have said at 55 amps discharge current you could run the load for about one hour if the battery has 100aH capacity. This is true for LiFePo4 batteries (which the OP has) but not lead acid batteries.

A lead acid battery with a rated capacity of 100aH is based on a much lower discharge rate. Typically, this is measured over 20 hours which for a 100aH battery would mean a discharge current of just 5 amps.

At higher currents the chemical changes in the battery become inefficient and the full capacity of the battery cannot be realised.

There is a thing called Peukert's law which can predict how long a battery will last under higher discharge currents. The Peukert constant for a battery reflects how capacity is reduced by higher discharge currents. The typical range of the Peukert constant for flooded lead acid batteries is 1.2 to 1.5. Feeding in the figures and assuming a Peukert constant of 1.2 gives just 67 minutes to fully discharge a 100aH battery at a 55A load. If the Peukert constant was 1.5 then the battery would be fully discharged in a little over half an hour. Or in other words you might, in the worst case, only be able to run a 600W load for about 15 minutes from a 100aH lead acid battery without going below 50% capacity, which is the recommended limit.

The Peukert constant for LiFePo4 is close to one which is why their capacity is hardly reduced by high current discharge.
 
With respect In reply to bigtwin this is why I work on 26 amp @ 12 volt d.c. for 1 amp @ 240 a.c. To allow for older batteries, undersized wiring, less efficient inverters. if you expect less and get more great but err on the side of caution.
 

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