## Battery usage time calculation

Gencis
Posts: 55
Joined: Fri May 04, 2018 2:45 am

### Battery usage time calculation

I need help in calculating battery usage time.

I have load (device) 1800W for 220V circuit

I am planning to use inverter 12V to 220V, nominal power 1500W/max 3000W

Battery 12V, 142Ah, 850A

1800W (load) / 12V (battery voltage) = 150A

So working time: 142Ah / 150A = 0,94h? and fully depleted battery?

JoeS
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Location: Silicon Valley, California

### Re: Battery usage time calculation

Gencis,

Before answering the question, let's have a look at your loads.

If I understand you correctly, you would like to use a 12v battery to power a 220vac inverter which will drive a 1800W steady-state load.

Your inverter is rated for 1500W but provides a 3000W short-term peak power which, for example, would allow for starting up a motor, but not for continuously driving anything at over 1500W. Thus, you have a problem as your inverter will be overloaded unless you find some way of reducing the power draw. For example, I have an 1800W induction cooktop that works off a sine-wave inverter just fine as long as I don't run it at full power.

Now, to your question -

For an output of 1800W if the inverter were 100% efficient and we run at a nominal 12v that would translate into 150A.

Inverter efficiency is unknown, but let's assume 80%. Thus your input 12v current draw is 187.5A.

Your battery capacity is 142Ah; however, battery capacity in ampere-hours is usually a 20-hour rating which means, in your case, 142Ah/20 = 7.1A can be drawn for 20 hours until the battery voltage drops to 10.5vdc.

Batteries are notoriously non-linear when it comes to available capacity vs. current draw (i.e., output power). The more current you draw, the lower the Ah rating.

For example, a very high-quality Deka Dominator Gel deep-cycle 4D battery has a 20-hr rating of 183Ah but a one-hour rating of 122Ah. In other words, one can draw 122Amps for only one hour from this 183Ah battery.

For argument's sake, let's use this ratio with your 142Ah battery: 142*122/183 = 95Ah is your battery's one-hour capacity; in other words, you should be able to draw 95A for an hour if it's a similar high-quality Gel battery.

But wait, you want to draw 187.5Amps! I have not found a capacity specification for less than one hour and the capacity continues to be non linear the more current you draw. If it were linear, you could assume roughly one-half hour.

At a full 1800W load, if the inverter works, I think you would be hard-pressed to get more than 15-20 minutes out of this battery before its voltage dropped to 10.5vdc, by which time the inverter would probably have turned itself off anyway.

Conclusion:

Realistically, ten minutes is about all you could probably hope for.

Edit
Addendum: the battery's CCA rating depends on whether it is SAEJ537, IEC, or DIN. Those tests are for very short-term starting-motor cold cranking capacity and don't apply to your inverter question. For example, the DIN CCA test takes your 850CCA battery frozen to -18°C and discharges it at 850A and to pass the test the voltage must stay above 9v for 30 seconds and 6v for 150 seconds.
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Don
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### Re: Battery usage time calculation

JoeS wrote:Batteries are notoriously non-linear when it comes to available capacity vs. current draw (i.e., output power). The more current you draw, the lower the Ah rating.

For example, a very high-quality Deka Dominator Gel deep-cycle 4D battery has a 20-hr rating of 183Ah but a one-hour rating of 122Ah. In other words, one can draw 122Amps for only one hour from this 183Ah battery.

Joe didn't mention it, but you really want to avoid pulling any lead acid 12 volt battery down to 10.5 volts as that damages your battery a good bit every time you do so and every subsequent discharge cycle will yield even less total energy than the previous one(s)

For large, continuous loads (like 1800 watts) you are much better off starting with a higher DC voltage. In this case, a pair of your 142 AH batteries in series, powering a 2KW, 24 volt inverter would be a much better solution. The DC current flowing in that circuit would be half of what it would be with a 12 volt inverter and less energy would be wasted as heat . . . . and you could run the load for more than twice as long, maybe getting you up to around half an hour

Don
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Gencis
Posts: 55
Joined: Fri May 04, 2018 2:45 am

### Re: Battery usage time calculation

Thank you very much for your help. I got a better understanding of batteries and how they work.

Don
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### Re: Battery usage time calculation

When you think of inverters and what it takes to power them, remember it's a ten to one ratio. 1800 watts at 120 volts AC is 15 amps and that same 1800 watts at 12 volts DC is 150 amps . . . . plus, there is usually a 10 to 15% loss in converting from 12 to 120, so it may be more like 165 or 170 amps. It takes a really powerful battery bank to put out that sort of power for any period of time and the inverter will need very large cables to combat the voltage drop. An inverter with a reasonable sized battery for power will put out enough power to run a microwave for 4 or 5 minutes, but if the load is to be sustained for a long period of time, things will begin to get warm and the battery voltage will start to drop fairly quickly

That said, there are big inverters which power RV air conditioners for several hours at a time, but they are almost always powered by lithium battery banks of several hundred amp hours - The Peukert Effect for Lithium batteries is much less pronounced than it is for lead acid batteries because they have a much lower internal resistance. That same low internal resistance enables them to be recharged much more quickly too. For high power applications, Lithiums have many advantages

Don
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