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Leisure battery - charging lipos advice please


Mowerman
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Posted by Shaunie on 02/01/2014 20:55:24:

My big point is that at the current we may draw for re-charging LiPos a lead acid will return only 50% of its claimed capacity and if we only wish to discharge to a 50% capacity then we will need four times the capacity of battery that our sums initially indicate.

Shaunie.

Shaunie, you have got me thinking about this and I have done a few simple sums:

With say a 120Ah battery that should be able to deliver 60Ah (50% of capacity), if we charge lipos on site for say 3 hours it would have to provide an average charging current of 20A. The capacity of batteries is specified at a given current and this is often 1/20 of the battery's capacity which in this example would be 120/20 = 6A. This is less than a third of the current I had hoped to draw from the battery. I suspect that at 20A the voltage of the battery would drop to a level at which the charger would no longer function.

It looks like site charging of flight lipos may be very demanding on batteries because it is over a relatively short period compared with say use in a caravan or boat.

Shaunie, your figure that a battery may need to be four times the capacity we need to take from it looks about right.

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Shaunie, your figure that a battery may need to be four times the capacity we need to take from it looks about right.

I think the important word is MAY, I would think a great many people field charge in this way with no problems, but then there will be those who charge from their car and need a bump start after wink.

I am only trying to make sure peoples eyes are open to the possibilities so that they can make an informed decision.

Shaunie.

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Warning, this post may damage your concentration. It’s just a load more gobbledegook.

As I mentioned in my previous post @ 01/01/2014 22:24:22, I’ve managed to find 5 3S 2,2Ah discharged packs and charge them, at a fairly sedate 1 amp rate.

My charger has an ‘always on‘ or idle current of 220 milliamps. As a comparison the Overlander RC-6S AC Pro has an idle current of 83 milliamps. The total taken from the supply battery here is 17.2Ah, the total input into the lipos amounts to 10.82Ah. However part of the charging was overnight, so I calculate that the charger was running on idle for about 10 hours, this was not part of the charging process so subtracting this from the total, 17. 2 - (0.22 x 10) = 15 Ah; this is the total used for charging. This makes the difference between in/out 15 - 10.82 = 4.18Ah. This is the best case situation, though. The input and output voltages are about the same, and at 2 amps charging rate, (2 outputs), the charger is still only dozing. If the charging rate was more aggressive, or higher cell number pack, this figure would undoubtably be higher. One way of making a simple calculation might be to multiply your pack(s) capacity by 1.5 or perhaps 2 and divide that into the supply battery capacity. This, though, always assumes the lipos are totally flat, and the supply battery is right up to the mark. If the lipos are not flat, that will help proceedings, but if the 12V is seriously off colour then that could curtail the flying time.

I’ve also found some test info on a cell, virtually all my paperwork went long out ago. This concerns a standard 100Ah cell, no idea of the manufacturer. Just to confuse the issue, the operator then would work in terms of the hourly rate, rather than the capacity. He’d also generally be more concerned about the time the battery is going to last, at a given rate of discharge; and, as Shaunie says, the higher the rate of discharge, the less capacity there is available. Also temperature makes a difference, if it starts to get a bit brass monkeyish then the capacity will also fall off a shade or more, too.

So, this battery is given a nominal capacity of 100Ah at the 5h rate; 5h is the normally accepted standard rate. Or when taking 5 hours to discharge, this amounts to 100/5 = 20 amp discharge rate. In terms of C this is only the reciprocal, 1/5 = 0.2C. The specification also includes the end of discharge voltage, here it’s 1.7V/cell, sometimes it’s 1.8V/cell that’s quoted; also it will state the ambient temperature, invariably 25 degrees C. Therefore we can basically discharge this battery at 20 amps and it will last for 5 hours, down to an end cell voltage of 1.7V. When going a bit faster, at the 3h rate, or 0.33C, 33A, then the battery is derated down to 90Ah, so it will run for 2.7 hours. The 2h, 50A rate = 76Ah capacity, and 1.52 hours. 1h, 100A = 56Ah, 0.56h, and the 30 minute rate, or 2C, 200A = 30Ah capacity, that’s less than a third of the original capacity. Now it will only run for 0.12h, or 9 minutes. But going the other way, a 10 hour discharge is rated at 120Ah. So when you know the current you want to supply, and the time you want it to be supplied for, you can calculate the necessary size of the battery.

Car batteries are rated a bit differently, the cold cranking amps is the current in amperes that the battery can supply for 30 seconds in a temperature of -18 degrees C down to a cell voltage 1.2V. The Reserve Capacity figure is the time in minutes the battery can supply a current of 25 amperes in a temperature of 25 degrees C down to an end cell voltage of 1.75V. So if the figures on the plate are say 400/70 this means it can supply 400 cold cranking amps or supply 25 amps at 25 degrees for 70 minutes.

Batteries are pretty boring old tack, really; trouble is, nothing much would work without them though, these days…….

PB

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