# Calculating Lead Acid Battery Self Discharge Rate

Discussion in 'Homework Help' started by JohnnyV, May 8, 2013.

1. ### JohnnyV Thread Starter New Member

May 8, 2013
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0
Hey All,

I've been a lurker for a while, finally found a problem I haven't been able to solve using the search function here.

I've been asked to find the self discharge rate in mA for a 12V lead acid battery. Here is what I know:
-The equation y=-0.001152*x+12.7 models the discharge of the battery (y = V, x = Days)
-The measured internal resistance of the battery is 3.649 mOhms
-A similar battery had a self discharge rate of 3.3 mA
-Assume constant temperature of 25 C

I've been struggling with the question itself - how does one have a "3.3 mA per day" discharge? Shouldn't this really be a value in Ah or watts? Is this sort of like saying you are losing 0.0792 Ah of capacity per day?

That being said, I've still tried, here is what I've thought about:

I know the measured internal resistance of the battery is 3.649 mOhm; My first thought was to calculate the voltage drop per day, then using Ohms law convert into A/Day then convert that is mA/day; this did not yield the results I expected (I got 548 mA). I know the answer should be around 3-5 mA.

My second path was to find the voltage drop over 180 days, calculate the power from that (P=V^2/R), then convert that into mA (A=sqrt(Watts/Ohm)); This did not yield what I expected either; I seemed to remember somewhere that Watts = Power/seconds, so I divided the power by 3600 (seconds in an hour), then converted to current. This yielded about what I expected (5.2 mA), but doesn't make sense to me. Is this pure chance or the correct way to calculate?

I've been mauling over this problem for around 2 days now, with little progress. I'm open to any comments, help, suggestions.

2. ### nsaspook AAC Fanatic!

Aug 27, 2009
2,826
1,970
The internal impedance model of a battery.

Rl is the internal load resistance of the cell. The value of Rl depends primarily on cell chemistry, state of charge, temperature. If the battery is new and the capacity discharge rate vs voltage is fairly linear then your voltage calculation should be a good approximation of the actual rate.

Life and Death: http://www.mpoweruk.com/life.htm

3. ### JohnnyV Thread Starter New Member

May 8, 2013
3
0
Hi nsaspook,

Not sure what you mean by "your voltage calculation should be a good approximation". I was given the equation to model the voltage. I'm trying to find the self discharge rate.

Thanks.

4. ### nsaspook AAC Fanatic!

Aug 27, 2009
2,826
1,970
My question would be how the change in voltage from self-discharge corresponds to the change in battery Ah capacity.
The self discharge rate in 'mA' for a 12V lead acid battery depends on it's Ah capacity for a known capacity loss rate (2% per month for a typical Gel cell). The internal series resistance limits the amount of current a battery can supply and still maintain output voltage but it doesn't directly correspond to true Ah capacity.

5. ### JohnnyV Thread Starter New Member

May 8, 2013
3
0
So the essence of the question is how to we find the capacity loss rate with the given information.